2212 Lecture notes - Chemistry & Physics

PHYS 2212

Notes follow and parts taken from Physics (6th Edition, Cutnell & Johnson) and Fundamentals of Physics (7th Edition, Halliday, Resnick, & Walker) Electricity & Charge All normal matter is made up of charged particles. “Charge” is just a way to measure how strongly two objects attract (or repel) one another electrically. Charge is similar to mass in that the product of the force between two charges depends on the sizes of those charges and is inversely proportional to the square of the distance separating them, just as we saw when we studied the law of gravity. Notable differences between the two situations include the possibility of repulsion between like charges (all masses attract one another gravitationally – even antimatter) and the vastly greater strength of the electric force. We observe charges to be quantized (i.e., not continuous, but having a smallest possible size, just as we now know water is not a continuous substance but a large number of molecules) in units of the charge on the electron, usually represented as e. The SI unit of charge is called the Coulomb (C) (we’ll define it later) and it is a very large number of charges. In fact, each electron has a charge of only -1.6 x 10-19 C. That means it would take a few billion billion of them to make one Coulomb. By convention, the charge on the electron is defined as negative, meaning the charge on the proton is positive. The signs are opposite, but the charges are exactly the same magnitude, to the highest accuracy we can measure. As always, there will be a few enhancements and corrections to the statements above that we’ll examine later. One is that experiments tell us that we can actually have charges which are 1/3 and 2/3 the size of the basic charge e when we look at the particles which make up protons and neutrons (called quarks). These have never been isolated individually, but their existence has been demonstrated indirectly. For this reason, we’ll still consider e to be the fundamental unit of charge. You’ve probably known since childhood that like charges repel and unlike charges attract. To this idea we’ll add the fact that charge is conserved. This law is absolute. No experiment has ever shown the overall destruction or creation of charge. We can separate charges very easily – rubbing a glass rod with a piece of fur, or your hair with a balloon will demonstrate this. These are methods of mechanically transferring electrons from one place to another. When you do this, the material that gave up the electrons becomes positively charged (having lost some amount of negative charge) and the material that picks up the electrons is negatively charged. The sum of the charges, though, is still zero (assuming the objects were neutral in the first place). Conductors & Insulators Just as we found last semester that certain materials conduct heat better than others, certain materials conduct electricity better than others (usually, we see that good conductors of heat are also good conductors of electricity). In a good conductor, the outermost electrons of the atoms (at least some of them) are not tightly bound. They are essentially free to roam around the conductor, and when they move, they carry charge from one place to another (generating a current, which we’ll talk much more about in a few chapters), or conduct electricity. Among the best conductors are gold, silver, copper, and aluminum (they’re also very shiny, and that’s not a coincidence). Things like quartz, glass, rubber, most plastics, wood, etc. are not good

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PHYS 2212 conductors, and are therefore called insulators. The atoms in these materials hold their electrons very tightly. Charging Materials If we put an excess of electrons on an object, we say that it has been charged. We could also take electrons from the object, and put a positive charge on it. If we want to try this on a conducting sphere, it needs to be insulated from the Earth. The reason is that the Earth is a huge conductor, and will gladly grab all of the excess charge it can reach. This is why we use lightning rods to direct lightning to the Earth – it can soak up the charge easily. We can charge the object by touching it with a charged rod and letting some of the charge transfer (it’s easier to say that than to say “letting electrons move from the object to the rod if the rod is positively charged” or “letting electrons move to the object from the rod if the rod is negatively charged”). The excess charge will quickly (very quickly) move to the surface of the metal sphere. It will do this because that’s the way it can get as far away as possible from the other charges of the same sign that came over with it from the rod. This is charging by contact. We can also charge by induction. Imagine connecting the metal sphere to the Earth (“grounding” it) and bringing a negatively charged rod nearby. The electrons in the sphere near the rod are pushed away, and the grounding wire gives them an exit to use. Some of them leave the sphere entirely, and will only return when the charged rod leaves. What if we break the grounding wire connection before letting the rod leave? Now we’ve driven electrons into the Earth, and they can’t get back to the sphere, so it is left with an overall negative charge. (Notice that it’s exactly the same size as the opposite charge we’ve added to the Earth, so we’re still not creating or destroying charge, just moving it). We couldn’t do the same trick with an insulating sphere, because the electrons near the rod aren’t free to move large distances away from their atoms, much less leave entirely. All they can do is shift position a small amount (small on the atomic scale) and spend more time on the far side of the atom they’re bound to. That will tend to make the insulating sphere’s surface slightly positive and give a small attraction between it and the rod. Coulomb’s Law Now that we know there is a force between all charges, we should find out the form of it. We’ll find that it looks almost exactly like Newton’s law of gravity, except we’ll replace mass with charge, since it’s generating the attractive force. As in the case of gravity, we’ll multiply by the inverse square of the distance between the two things, and we’ll also need a constant to make the units balance. This one is sometimes written as k and its value in the SI system is 8.99 x 109 N m2/C2 (notice the similarity in the units between this and “G” – we replace kg by C, but that’s it). Notice that while G is small, k is very large. This reflects the difference in strengths of the two forces. For reasons we’ll look at later, we sometimes write k in a different form: k = 1 / ( 4 π ε 0 ) where ε0 is called the permittivity of free space and equals 8.85 x 10-12 C2 / (N m2). Therefore, we write Coulomb’s law as

F=

k q1 q 2 q1 q 2 = r2 4π ε 0 r 2 2

PHYS 2212

where q1 and q2 are the sizes of the two charges. If they are the same sign, we’ll get a positive force (repulsion). If they’re opposite, we’ll get a negative force (attraction). This force is, like all forces, a vector. That means it has magnitude and direction. The direction will be either towards the other charge (attraction) or away from it (repulsion). In one sense, now that we know the form for the force, our work with Coulomb’s law is done. It’s now just one more force we need to consider when determining the motion of an object. It’s part of Newton’s 2nd law as a force to be included in

∑F

=ma

(x direction, with a similar force in the y x x and z directions). We’ll find its components just as we’ve done with other forces. If we want to know the effect of several charges on another charge, we’ll use Coulomb’s law between the charge we’re interested in and each of the others to find the forces involved, then we’ll get components, add them up, etc. There’s nothing particularly special about this new force. For fun, let’s find the gravitational force between two protons and the electrostatic force between them. The charges are both 1.6 x 10-19 C, the masses are both 1.67 x 10-27 kg, and we’ll put them 10-15 m apart (i.e., next to each other in the same nucleus). We get:

FGrav =

G m1 m 2 2

( 6.67 × 10 =

−11

)(

(

r 10 while for the electrostatic case, we have:

(

)(

N m2 / kg 2 1.67 ×10 −27 kg 1.67 × 10 −27 kg −15

m

)

2

)(

)(

) = 1.86 ×10

− 34

N

)

k q1 q2 8.99 ×109 N m 2 / C 2 1.6 ×10 −19 C 1.6 ×10 −19 C FElec = = = 230 N 2 −15 r2 10 m

(

)

Look at the difference! The electrostatic force is 1036 times larger! That’s a trillion trillion trillion times. This is roughly the same size difference as the diameter of an atom compared to the diameter of the entire universe! If charges move under the influence of this (or any) force, they will create an electric current. Current is just the rate at which the charge in a region changes. In an infinitesimally small time, we will see that the current I is found from

I=

dQ dt

The units of current are Coulombs/second, or amperes.

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PHYS 2212 Conductors and Shielding If there is a net charge on a conductor (too many or too few electrons), the charges will all go to the surface in an effort to get as far away from each other as possible, since like charges repel and conductors allow charges to move freely. What does that mean for the electric field inside the conductor? You can answer this by thinking about it either of two ways: first, if there was an electric field, the charges in the conductor would move in response to it. The electrons would move in a direction opposite to the electric field, and each one moving that way would act to reduce the electric field, until the field was zero (this happens in a very tiny fraction of a second in a conductor – essentially instantly). The other way you can see the field must be zero is to imagine the charges on a conducting sphere. They will be spread evenly along the surface, with no particular point singled out. How would the electric field know which way to point? If there is a cavity inside the conductor, there will be no electric field inside it. This provides a way to shield a region from external electric fields. This kind of sheltered environment is called a Faraday cage. If we put a conductor into an external electric field, it will alter the field slightly. Wherever the field touches the conductor, it will be perpendicular to its surface. It has to be – a component along the surface would push electrons around in the conductor until they built up and cancelled that part of the field. Everything we’ve talked about depends on the free movement of charges – if we’re in an insulator, we can’t assume any of this is true. The Electric Field Sometimes, the concept of a field is a useful one. We’ve already used this idea when talking about the weight of an object on Earth’s surface. What we’re really doing is finding the force of gravity between some small mass and the rest of the Earth. Since, in so many cases, we know that the separation between the centers is essentially just the Earth’s radius, we can do some of the math in advance and multiply the Earth’s mass by G and divide by the square of Earth’s radius. After doing that, if we want the weight of an object on Earth’s surface, we just multiply it by the number we found above (which we call “g”). This g is sometimes called the gravitational field of the Earth. All we need to do to turn the field into a force is to multiply by the mass we’re interested in. Since we know force is a vector, and mass is a scalar, the field must be a vector (pointing into the Earth). We can use the same idea to find the electric field produced by a (large) charge Q. Now it will be force divided by charge rather than mass:

  F E= q0 where q0 is the magnitude of a (small) test charge, F is the magnitude of the force between Q and q0, and E is the electric field produced by Q. We’ll use the letter E for the electric field, and measure it in Newtons per Coulomb (i.e., if the field strength at a point is 25 N/C and we place a charge of 10-15 C at that point, the force on it should be 25 x 10-15 N). Charge is a scalar, so this field is must be a vector. We’ll define it as pointing from positive to negative charges. In this picture, a proton will look like a pincushion with electric field lines streaming out of it in all directions. Electrons will look the same, but the lines will be pointing inward. We define it this 4

PHYS 2212 way so that the force will equal the charge times E. For another electron, the charge will be negative, and therefore E must be negative also to get a repulsion. More lines will mean a stronger field.

Electric field lines

Isolated positive charge

Isolated negative charge

The restriction on this idea is that we can detect the magnitude & direction of the electric field by measuring the force on a charge as we move it around in the field, but we need to use a small charge. If we’re trying to find the field of a 10-17 C charge, we won’t be able to do that very easily if we’re using a 10 C charge to do it! The small charge’s field will become completely overwhelmed and dominated by the field of the large charge. For this reason, we will usually talk about the motion of a test charge in the electric field. This is a charge which is tiny on the scale of the charge producing the field we want to measure. We do this so that we can get a more general result than Coulomb’s law alone would give us. We avoid many long calculations for the gravitational force by defining g, and we’ll avoid a similar amount of work by defining E. Just as g is independent of the mass we’re measuring, E is independent of the magnitude or sign of the test charge we use. For a point charge, then, the E field looks like this:

E=

q 4π ε 0 r 2

Notice that this field depends on distance; this shouldn’t be surprising since the force between two charges itself depends on distance. Other situations produce other fields. What are the rules for drawing field lines? First, we know they go from positive to negative charges. Because of that, they can’t stop in empty space. Also, since they represent the path a positive test charge q would follow, they can’t cross (which way would the charge go at the intersection?). For larger charges, we should draw more lines and put them closer together. If we put a positive charge near a negative charge, they will be connected by many field lines, and many others will go out and terminate on (or start on) far away charges. Your book has a picture of this configuration, known as a dipole (because you have two opposite poles near each other). If the two charges in a dipole are the same magnitude and opposite sign, the total charge in that region is zero. This doesn’t mean that the field around the charges is zero

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PHYS 2212 everywhere, though. The reason is that a point will be equally far away from both charges only on a plane between the two charges, and even there, the field won’t cancel because of the different signs. As you get far away from the dipole, the field gets small (faster than the field of a single point charge). Where the field of a single charge falls off as the inverse square of your distance to it, a dipole field drops off as the inverse cube of the distance. The reason is that, as you get farther away from the two charges, the difference between your distance to the positive charge and your distance to the negative charge approaches zero. If the distances were exactly the same, the charges would cancel and we’d have no field!

Notice that at every point, the electric field is tangent to the field lines. We can talk about the strength of the dipole by multiplying the size of one of the charges (since they’re equal) by the separation between them. This strength, called the dipole moment, is written as

p=qd If we move a distance much greater than d away from the dipole, the electric field can be calculated from

E=

p 2π ε0 z 3

6

PHYS 2212

Electric Fields from Continuous Objects For macroscopic collections of charge, we are able to ignore the fact that charge is quantized and treat it as a continuous “fluid”. The electric field produced by this extended array of charges is found by adding up the contribution from each tiny piece of it. To do this, we need calculus. We define charge densities in one, two, and three dimensions. A line charge will have a linear charge density (Coulombs/meter), a sheet of charge will have a surface charge density (C/m2), and a volume of charge will have a density measured in (C/m3). In every case, we’ll find the electric field by doing some form of the integral below:

dq 4π ε 0 r 2

E = ∫ dE = ∫

For a line charge, we’ll let the linear charge density be λ and write dq = λ ds where ds is an infinitesimal length of the line of charge. To make the problem simpler, assume the line charge is bent into a circle of radius R and we want to know the electric field along the axis of symmetry of the circle at a distance z above it. Pieces of the circle on opposite sides will partially cancel each other’s electric fields. z

dE

R

The only part of the field that doesn’t cancel is the part perpendicular to the plane of the circle (make this the z axis). To find that component, we have to multiply by the cosine of the angle between the electric field direction and the z axis. This can be written as

z z cos θ = = r z 2 + R2 Performing the integral above, which stretches around the circle’s perimeter, we get

 2π R 2π R E = ∫ dE cos θ = ∫ 0

0

z λ ds

(

4π ε 0 z 2 + R 2

or, writing the total charge on the ring as q, we get

 E=

(

qz

4π ε0 z 2 + R 2

7

)

3/ 2

)

3/ 2

PHYS 2212 Notice that very far from the ring (z >>R), this will look just like the electric field of a point charge. We can turn this circle into a disk carrying a surface charge and see what that does to the electric field (again, along the z axis). The only big change is that the charge element dq goes from being λ ds to σ dA. For a circle, dA is just r dr dθ, but this problem is symmetric about the z axis so we can do the θ integration immediately and write dq = σ dA = σ (2 π r dr). We get

dE =

σz 4ε 0

2 r dr

R

∫ (z 0

2

+ r2

)

3/ 2

 σ  1 − ⇒ E= 2 ε 0 

  2 2  z +R  z

We can find the field from an infinite sheet of charge (which is sometimes a useful idealization of a real situation) by allowing looking at the limit of the above expression as R  , so

 σ E= 2ε 0 While there’s no such thing as an infinite sheet, what’s really important is that we get approximately the same result when R >> z, which is a possible (and not uncommon) situation. Charges in Electric Fields It’s not too surprising that a single point charge should move when it finds itself in an electric field. Our definition of the field was just the force experienced by a charge divided by the size of that charge. What is also interesting and a little more complex is the behavior of a dipole in an electric field. If the electric field is uniform, the force on each charge in the dipole will be equal and opposite. As you’ve seen in the first semester of physics, if an object is acted on by two equal and opposite forces, its center of mass won’t move. However, if those two forces are not acting on a line which goes through the center of mass, they will generate a torque. One end of the dipole will move “downstream” along the field lines, and the other end will move “upstream”. This will cause the dipole to align itself with the electric field. You’ve seen (basically) the same effect if you’ve ever brought one bar magnet towards another; the magnet you approach will rotate around until its poles are along the same line as the approaching magnet. In the uniform field, the force on each charge will have a magnitude of qE and each will generate a torque (in the same direction – that’s why the dipole rotates) of qE multiplied by the distance to the center of mass (d/2) multiplied by the angle between the force & displacement (sin θ). Added together, we get

8

PHYS 2212

    d  τ =  q E sin θ  × 2 = qEd sin θ = p × E 2   It’s worth mentioning how important the uniformity of the field is; if the field seen by the two charges is different, the forces won’t cancel and the dipole itself will move. A point charge has a field which is non-uniform everywhere. A dipole near an isolated positive charge will rotate so that the negative end of the dipole is closer to the positive charge; at that point, the forces won’t balance and the dipole will move towards the isolated charge. Uniform fields are another example of a useful idealization of reality. Last semester, you probably considered the Earth’s gravitational field to be uniform for the first few days you studied it. To do this, you just set the acceleration caused by gravity to be equal to g and assume that it’s the same everywhere. This is a very powerful approximation, since it lets you quickly find the maximum speed of a falling object or its potential energy, or the weight of something on the Earth, etc. You probably also learned that you can’t use g = 9.8 m/s2 to calculate the gravitational pull of the Earth on the Moon. The reason is that the real formula for the Earth’s gravitational field is

 G M Earth g= r2 When you’re looking at things as small as people and skyscrapers, the difference between using Earth’s radius as r and Earth’s radius + the object’s height as r is negligible. As r gets noticeably larger that the Earth’s radius, the discrepancy grows. The fact is, Earth’s gravitational field is really non-uniform everywhere, and the same principle (applied to the Moon and Sun) gives us the main cause of tides! Why is this more of an issue this semester than last? It’s because of the relative strengths of the electromagnetic & gravitational forces. Only very large collections of mass, like the Earth, have strong enough gravitational fields to be of interest to us at this level. Large collections of mass typically have large (on the human scale) sizes – many km or hundreds of km or more. Therefore, moving things tens or hundreds of meters (human-scale units) won’t reveal large changes in g. The electromagnetic force, on the other hand, is much stronger and even tiny collections of charge (physical sizes much less than 1 meter) can produce easily measurable electric fields. When we move meters or hundreds of meters from this tiny collection of charge, we’ve changed the size of the electric field we measure by a large amount. Back to the dipole. If it will rotate under the influence of a field, there must be a potential energy involved. The potential energy will be

U = −W = −

θ

∫ τ dθ = − p E cosθ

−90

o

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PHYS 2212 if we set the zero of potential energy to be the point where the dipole is aligned perpendicular to the field. Gauss’ Law Finding the electric field of something other than a point charge can be a difficult problem. Fortunately, if we have a situation of high symmetry, there is a technique we can use to quickly find the electric field’s magnitude & direction. To use this technique (known as Gauss’ law) we need to introduce the idea of flux. This can be thought of as the number of electric field lines entering (or leaving) a certain area in space. To simplify this idea, we’ll use an ordinary square region as our area. You can see two different situations below; in each case, the strength of the electric field (density of field lines) is the same, and the area of the square is the same.

The first diagram shows the square oriented so that the maximum number of field lines pass through it; this makes the flux a maximum. In the second picture, no field lines go through the square, so the flux is zero. Flux is therefore a kind of projection, and there will be a Cos θ term in the formula for it. It’s not hard to see that a larger square (more area) would give a larger value of flux, and a stronger electric field (more dense field lines) would also give a larger flux. The formula is then seen to be

Φ = E A Cosθ where Φ is the flux, E is the electric field, A is the area involved, and θ is the angle between the electric field and the vector perpendicular to the area. More generally, we’ll want to know the flux through a closed surface (this will be the boundary for some volume; the six square faces of a cube form its boundary, the 4 π r2 surface area of a sphere forms a boundary for its volume, etc.) If the surface is arbitrary, we have to do an integral to find the flux since A and E won’t (in general) be constant. Now we can write

  Φ = ∫ E ⋅ dA Why do we care about flux? The same reason we care about any quantity like this in physics – it’s conserved, or related to something that’s conserved. In this case, the flux is proportional to

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PHYS 2212 the charge enclosed by our closed surface defining the region of integration above. This is Gauss’ law, and we can write it as

  ε 0 Φ = ε 0 ∫ E ⋅ dA = Qenclosed This is an incredibly powerful law. For example, if I want to know the flux through a region of space shaped like a cow, all I really have to know is how much charge (net, of course) is contained in that cow-shaped volume of space. If it’s zero, I know the flux is zero. If I bring a large positive point charge near (but not inside) that same volume of space, lines of the E field will be entering one side of the cow and exiting the other. The density of lines will be greater on the side of the cow nearest the charge. Different tiny patches of area on the cow will be pointing in different directions relative to the electric field passing through the cow, which will also point in different directions at different parts of the cow. This would be a horrible calculus problem (and a horrible cowculus problem). However, if we know there is no charge inside the cow, we know that E and each little piece of cow skin dA must be arranged in such a way as to cancel out when the contributions over the whole cow are added. In the drawing below, if the surface on which we apply Gauss’ law is the red dashed sphere surrounding the positive charge, we can see all sorts of arrows leaving it – electric field is flowing out of it, and there is no balancing electric field flowing in. If our surface is instead the blue dotted sphere, as much electric field is flowing out of the right side as is entering the left side. The net flux is zero, because the enclosed charge is zero. (If the charge below were negative, everything would look exactly the same except for the direction of the arrows, which would be reversed).

Gauss’ law ↔ Coulomb’s law Both Gauss’ law & Coulomb’s law talk about the forces or fields (which we know are tightly connected) produced by and acting on charges. To connect them directly, imagine that our closed surface in the integral mentioned earlier is the surface of a sphere with a charge q at its center. If we give the sphere a radius r and assume (for simplicity) that q is positive, we will have an electric field directed radially outward at all points on the sphere’s surface. Since the charge is at the center, all points on the surface are the same distance (r) from the charge, so the electric field has the same magnitude everywhere on the sphere. Also, the normal to a tiny piece of the surface 11

PHYS 2212 of the sphere will point radially outward, so the cosine term will be one everywhere. The only non-constant thing left to integrate is the area of the sphere, so we get

ε 0 E ∫ dA = q The area of a sphere is just 4 π r2, so we can rewrite the equation above two ways:

(

q =ε 0 E 4π r 2

)

or E =

q 4π ε 0 r 2

Charged Conductors What will happen if we pile charges up on a conductor which is insulated from the Earth so that they can’t leave? How will they arrange themselves? We know that like charges repel each other, so they should try to get as far from one another as possible. It might seem that the best way to do that is for the charges to distribute themselves uniformly throughout the conductor so that each one has the largest possible volume to itself. In fact, the charges will all collect on the surface, since this will minimize their energy. Gauss’ law can be used to demonstrate that the charges have to appear on the surface. The main idea here is that we can’t have an electric field inside a conductor and expect it to last. The defining characteristic of a conductor is that charges are free to move in it. If we found a way to apply an electric field inside the conductor, the charges there would just rearrange themselves until their field cancelled out the applied field. (There are situations where electric fields can exist in a conductor, but we would then be looking at electrodynamics instead of electrostatics). Using this idea, imagine that the closed surface in the integral form of Gauss’ law (known as a Gaussian surface) is one that is just barely underneath the physical surface of a conductor. Whether the shape of the Gaussian surface is the same as the shape of the conductor, or some other weird shape that will also fit inside the conductor, the E field has to be zero, or the charges will just rearrange themselves until the field is zero. This means that E is zero regardless of our choice of surface, so the flux must be zero also. If the flux is zero, the charge enclosed by the Gaussian surface is also zero. The only place left for the charge to be is on the physical surface of the conductor. What if the Gaussian surface is larger than the conductor? In the case of a sphere, the charge will be uniformly distributed on the surface. If the surface isn’t spherical, the charge distribution will be more complicated. We can simplify the situation by breaking the surface into many tiny regions, each of which will appear to be flat, just like the surface of the Earth appears to be flat on the small scale. To make the integral easier, our Gaussian surface will be a cylinder with one end inside the conductor and the other end outside. The axis of the cylinder is parallel to the normal of our small piece of the surface. The E field will also have to be normal to the surface, because any component of E parallel to the surface would cause charges to move around until

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PHYS 2212 that component was neutralized. So, looking at the surface of our cylinder, we see that 1) for the inner face of the cylinder, E =0 so there is no contribution to the flux 2) for the “length” of the cylinder, E is perpendicular to that part of the Gaussian surface, so there is no contribution to the flux. For the final part of the cylinder, though, we’ll have a circular end piece whose normal is either parallel to E (if the field lines are leaving the surface = the conductor is positively charged) or anti-parallel to E (negatively charged conductor). We’ll get that

E=

q enc σ = Aε 0 ε 0

where σ is known as the surface charge density and will be measured in C/m2. This will be give us the electric field for any small segment of a conductor (although we have to keep in mind that, for irregularly shaped conductors, σ itself will vary over the surface of the conductor as charge piles up in some places and avoids others). Applications of Gauss’ Law Gauss’ law is most useful when we pick a Gaussian surface with the same symmetry as the underlying charge distribution. For example, if we have a line charge, we should make the Gaussian surface a cylinder. If this line is infinitely long (not as ridiculous as it sounds – if your distance to a long, thin charged line is small compared to the line’s length, this is a good approximation), the electric field can’t have a component along the line’s length. This means the cylinder’s end pieces won’t contribute to the flux at all. The area of the rest of the cylinder would just be 2 π r h, and the cosine term will be one. That means Gauss’ law will look like

ε 0 E ( 2 π r h ) = qenc so E =

λ 2π ε 0 r

where λ is the linear charge density (measured in C/m). For an insulating, infinite sheet of charge, the electric field will again have no component in the plane of the sheet by symmetry (which way would it point?). If the electric field is normal to the surface (on both sides), we can again use a cylinder as our Gaussian surface, with the axis of the cylinder parallel to the normal and the sheet passing through the middle of the cylinder. Now there is no electric field through the main body of the cylinder (since the cosine term will be zero here) and any contribution will have to come from the circular end pieces (each of area A). Gauss’ law gives

ε 0 ( EA + EA) = qenc so E = 13

σ 2ε 0

PHYS 2212 If instead the sheet is conducting, we can make a useful device (known as a capacitor – more about that later) by bringing a positively charged sheet near a negatively charged sheet. Since the two sheets are conducting, the charges on each will be free to move to the side nearest the other (oppositely charged and therefore attractive) sheet. The field we get in this case is uniform in the limit of infinitely large sheets. The fact that the charges are mobile changes things compared to the insulating-sheet case. One of these conducting sheets by itself would have its charge spread over both faces to give some net surface charge density σ1 on each side with the same electric field on each side. When the two oppositely charged conducting sheets are brought near each other, all the charge on each plate moves to the inside surface, leaving nothing on the outside. This gives a surface charge density twice as large on the inside faces (2 σ1) and no surface charge on the outside faces (meaning no electric field outside of the plates). In effect, the charge (and electric field) that used to be on the “outside” of a plate gets moved to the inside, adding to the charge and field there. Finally, what about a spherical arrangement of charge? The book mentioned earlier that 1) if a charged test particle is outside of a shell of uniform charge, the shell acts like a point charge as far as the test particle is concerned and 2) a charged test particle inside a shell of charge will feel no force on it at all. You can prove this by imagining a spherical Gaussian surface either inside or outside of a charged shell. If the surface is inside, the enclosed charge is zero and the electric field is therefore zero (meaning no force on a charged test particle). If the surface is outside (a distance r from the center), the electric field will just be the enclosed charge divided by the surface area of the Gaussian sphere. This would give an E field of

q enc E= 4 πε 0 r 2 which you should recognize as the electric field at a distance r away from a point charge of magnitude qenc. Potential Energy In the same way that a mass in a gravitational field has potential energy equal to mgh, a charge in an electric field has a potential energy. The mass would “like” to move in the direction of the gravitational field, and the charge would like to move either in the direction of the field (positive charge) or opposite to it (negative charge). Both gravity and the electric force are conservative (remember that that means a decrease in potential energy is matched by an increase in kinetic energy), which is why we can define a potential. As before, work is force multiplied by distance. In the case of gravity, that gave us mgh (near the Earth’s surface, at least, where we can say that the gravitational field is a constant g). If the electric field is constant, we’ll just get the expression below

W AB = q E ( a − b ) 14

PHYS 2212

where a and b are the initial and final locations of the charge q. As we’d expect, the work done depends on the size of the charge, just as the work done by gravity depends on the mass which is moving from place to place. We’d like to eliminate the dependence on the test charge. We do that by defining an electric potential which is related to, but not the same as the electric potential energy. All we have to do is divide both sides of the equation above by the charge q. This will prove to be so useful that we will give this new quantity (work divided by charge) a name of its own. We’ll call it V for voltage and the units will be J/C, or volts. Part of the value of this will be that we’ll automatically know the change in a charged particle’s energy when it goes between two points of different voltage if we know its charge. For this reason, we’ll sometimes refer to voltage as the potential difference. We can also write the units for electric field strength in terms of voltage by using 1 N/C = 1 V/m. This is sometimes more convenient; for example, the potential difference between the two terminals of a car battery is about 14 volts and the terminals are typically around 0.3 m apart. That gives an electric field between the terminals of 14 V/ 0.3 m or about 46 V/m. An electron moving from the negative pole of a battery (AA, AAA, C, or D) is accelerated through a potential difference of 1.5 volts. That means that each electron will experience an increase in energy of

W = qV = (1.6 ×10 −19 C ) (1.5V ) = 2.4 ×10 −19 J This is a tiny amount of energy, of course. A more convenient unit of energy when we’re actually interested in very tiny things like electrons is the electron volt. This is easier because an electron moving through the 1.5 V potential difference would have a change in energy of 1.5 electron volts. A little bit of math will show that 1 electron volt = 1.6 x 10-19 J. If we want, we can find the increase in an electron’s velocity when it goes through a given potential by assuming that the potential energy becomes kinetic energy. While electrons generally do the moving when there is a potential difference applied to something, it was originally believed that positive charges were moving to cause the electric current. Electrons are said to have a negative charge, though, so the flow of electrons is from a negative battery terminal to a positive battery terminal. When we look at currents, we’ll generally refer to them as moving through a circuit from the positive terminal to the negative terminal, even though we know that’s not exactly what’s happening. Equipotential Surfaces and Electric Fields Just as we saw with the gravitational field, a charge will only do work (or have work done on it) when it gets closer to or further away from another charge. When we move a mass around on the ground (assuming g is constant and the Earth is a perfect sphere), we don’t do any work against gravity. If we move one charge around another charge, keeping the distance between the two constant (tracing out the surface of a sphere), we won’t do any work. From the formula above for potential, every point on the sphere’s surface will have the same value of V. For that reason, we call it an equipotential surface. We can move a charge anywhere on an equipotential surface without doing any work. You’ve seen this idea before – on a topographic map, contour 15

PHYS 2212 lines connect points of equal height. You could move a mass around this line without doing any work. Also, you see the same thing on weather maps – contour lines connect points of equal pressure (isobars). When lines like this are closely spaced, that means there is a more dramatic change in what they measure than in a region where the lines are far from one another. When the lines on the weather map are closely spaced near you, there will be high winds (driven by a dramatic change in pressure over a small distance). When there are closely spaced lines on a topographic map, you have a very steep incline to go up (or down). What do we get when lines of equipotential are closely spaced? A large electric field! We can define the electric field as the something called the gradient of the electric potential. The gradient is a measure of the direction of the most rapid change of something. If you’re on a hill and set a bowling ball down, it will roll in the direction where the height change is most rapid. If we let go of a charge at some point, it will move in the direction where the electric field is largest (so the potential is changing most rapidly). Since negative charges move in the opposite direction that the electric field points, we need to define the electric field like this:

E=−

dV ds

where dV is the change in potential between two points and ds is the distance between the two points. Inverting this allows us to find the potential from the field:

f

  V f − V i = − ∫ E ⋅ ds

i The potential difference is always taken between two points, but we may set the potential at one point to be zero. This is exactly what we do with the gravitational field; if you want to know how fast something is moving just before it hits the ground, you need to know its starting point and its endpoint. Sometimes it is convenient to make sea level the zero of height, but if we were looking at something falling from a desk on the 80th floor of a skyscraper to the carpet in the same office, we’d probably measure the height of the desk from the floor rather than sea level. No matter how we choose our zero point, the difference between initial and final points is all that matters.

16

PHYS 2212

In the drawings above, we have the electric field lines on the left and the equipotential surfaces on the right. The electric field is much stronger between the two charges where the equipotential surfaces are close together than far away from the charges. Notice that the electric field is perpendicular to the equipotential surfaces everywhere. For this reason, also notice that we never see equipotential surfaces cross (since each represents a different potential, what potential would we have at the crossing? And what way would a charge move at that point?). In a conductor, all points are at the same potential (so it’s kind of an equipotential volume rather than surface). The reason is the same thing that makes it a conductor – the charges inside it are free to move in response to an electric field. If there were a field of some kind in the conductor, the charges would move in response to it. Once enough electrons moved in the direction opposite to the field, they would create a field of their own which would cancel out the original field. It’s the same idea as assuming all points on the surface of the water in a bucket are at the same height – if there were a small hill or valley in the water, it would very quickly be erased by the movement of the rest of the water. Potential Difference created by a Point Charge We shouldn’t be very surprised that a point charge produces its own electric potential as well as responding to the potential in its area. We generally ignored it last semester, but a mass moving in the gravitational field of a planet will also create its own gravitational field. For both masses and charges, something that can sense the field (and move in response to it) will also create its own field. Before we define this potential difference, we should recall something from our discussions of gravitational potential energy. We claimed that it was just mgh. We know, though, that work is force times distance. What we’re saying when we use mgh is that the force (mg) is independent of distance, so we can find work by simple multiplication. If h is very small when compared to

17

PHYS 2212 the Earth’s radius, this approximation will be fine. However, it is relatively common to work with electric fields which can’t be considered constant over the distances that interest us, and we can no longer just multiply a single value for the field by the distance a charge is moved in that field to find potential energy. As you might guess, correctly finding the potential energy in this case will require calculus. We need to find the area under a graph of force vs. distance, and that will give us work (or energy). Another similarity between the electric force and gravity is that motion around the charge or mass is “free”. Just as the potential energy doesn’t change when we move a mass along the ground at a constant height, we won’t see the electric potential energy change if we move around a charge without changing our distance to it. If we’re a distance R away from the point charge, the integral we need to do is ∞

q V f − Vi = − ∫ E dr = − 4π ε 0 R

∞

dr ∫R r 2

It’s common to set the potential to zero at infinity, so we will be left with

V=

q 4π ε0 r

Notice that the potential is a scalar. We don’t have to worry about components here. This can be a real simplification when we want to see how a test charge moves under the influence of lots of other charges. We don’t have to find the vector components of the electric field from each charge and then combine them – we can just add the potential difference produced by each one at the location of our test charge to find the total potential difference. The potential produced by a charge will have the same sign as the charge. If we have a positive charge and a negative charge near each other, there will be a place on the line connecting them where the potential is zero. Keep in mind that that doesn’t mean a charge placed there would stay at rest – the direction of movement depends on the way the potential changes with position and the sign of the test charge. This means that the general formula for the potential caused by a group of N point charges can be written as

1 V= 4π ε 0

18

qi ∑r i =1 i N

PHYS 2212

Dipoles We’ve already looked at the electric field produced by a dipole – what about the potential around one? We can use the formula above and write it as

q Vdipole = 4π ε0

1 1  −   r+ r− 

where r+ and r- just represent the distance between the + or – charge and the point where we’re measuring the potential. If the distance to the point of measurement is large compared to the separation d between the two charges, the term in parenthesis above is approximately equal to (d Cos θ / r2) where the angle θ is measured from the plane that is normal to the dipole axis and centered between the two charges. This gives

Vdipole =

p Cosθ 4π ε0 r 2

A dipole moment p can be induced in a collection of point charges by an applied field. Once that happens, the dipole can respond to the field (notice that we might have predicted that an electric field will have no effect on, say, a neutral hydrogen atom composed of one negative electron and one positive proton. We’d have been wrong!). In fact, molecules can induce dipole moments in each other and then respond to the induced moment. This is the source of the van der Waals force that you may have heard about in chemistry class. Potentials & Continuous Charge Distributions The general procedure to find the potential produced by a continuous charge distribution is really just an extension of what you’ve seen above. We let the sum over all charges become an integral and the size of each charge becomes infinitesimal. We can then write

V=

1 4π ε 0

dq ∫r

Your book works out the formula for the potentials due to line charges and charged disks. Rather than reproduce that here, we just note that we could predict the general form for these things using what we’ve already found about their electric fields. The electric field of the line charge was found to be a constant multiplied by (λ/r) where λ was the linear charge density and r was 19

PHYS 2212 the distance from the line to the point of interest. Since potential is basically the integral of the electric field, and we know that the integral of (dr/r) is the natural logarithm of r, the result found in your book is no surprise. Similarly, we found earlier that the electric field due to a charged disk was a constant multiplied by

 z σ 1 − z 2 + R2 

   

and, if we find the integral of that with respect to z, we will get a term proportional to

z − z2 + R2 as you can see in your book. Field from Potential Once the potential is known in an area, the electric field is relatively simple to find. It points from positive charges towards negative charges (higher to lower potential) and is perpendicular to the equipotential surfaces. There must be some directional dependence, since we’re finding a vector quantity from a scalar. The component of E in a given direction is just the negative of the rate of change of V in that direction, so we can write

Ex = −

∂V ∂x

Ey = −

∂V ∂y

Ez = −

∂V ∂z

Electric Potential Energy A charge in an electric potential has a potential energy just as a mass in a gravitational potential does. That potential energy is just the product of the charge and the potential. If the potential is due to another charge, we can write the potential energy of the system as

U=

q1 q 2 4π ε 0 r

20

PHYS 2212

Notice that the sign of the energy works out automatically. If the two charges have opposite signs, we’ll get a negative value of potential energy, meaning we have to put energy into the system to separate the charges. If the signs are the same, we have a positive potential energy, meaning repulsion. If there are more than two charges, the total potential energy is just the sum of the potential energies found using the formula above for each possible distinct pair of charges. Capacitors and Dielectrics If we arrange two metal plates so that they are facing each other, we can use this setup to separate and store charge. We’ll have positive charges on one plate, and negative on the other. This is called a parallel-plate capacitor. Capacitance is basically the ability to store charge. When we pile these opposite charges on the plates, an electric field will develop (pointing from the positively-charged plate to the negatively charged one). The field lines will be perpendicular to the plates, so that our configuration looks like this:

+ + + + + + +

-

Using a method we’ll examine shortly, we can find that the electric field inside the capacitor is given by

E=

σ ε 0 A ε0 q

=

where q is the total charge on each plate and A is the area of each plate. Interestingly, the strength of this field doesn’t depend on the position between the plates – it’s no larger close to one plate than in the center. In the real world, the field will bulge out a little near the edges of the plate. In this arrangement, each plate will have a charge of the same magnitude but opposite sign. The formula we use to determine the amount of charge on each plate is just

Q=CV where Q is the charge per plate, V is the voltage of the battery, and C is called the capacitance. This factor contains information about the geometry and construction of the capacitor, and the 21

PHYS 2212 units of capacitance are called farads (1 F = 1 Coulomb/Volt). Farads are huge units of capacitance, so we’ll typically use things like microfarads, nanofarads, and picofarads. As C gets larger, the amount of charge a battery of a given voltage can separate increases. An electric field will be developed between the plates of the parallel-plate capacitor, and it is (approximately) uniform. Its magnitude will be E = V/d where d is the distance between the plates. At the edges of the plates, the field will get a little strange, but we’ll generally ignore that. The electric field will point from the positive plate to the negative one, as always. The formula for C for parallel-plate capacitors (derived in your book) is just

C=

κ ε0 A d

We’ll talk about what κ is later. If we’re in a vacuum, it’s just 1. Capacitors can also be made in other shapes, and the calculus used to find the capacitance of these shapes is shown in your book. The general procedure to find it uses Gauss’ law to find the electric field and the fact that the potential difference is the integral of the electric field over some path. You should notice that capacitance calculations always leave you with a term proportional to ε0 multiplied by a length. Capacitors in Series and Parallel We can combine capacitors in different ways to change the total capacitance of the collection. This is very valuable, as we can have just a few different capacitors and use them to make almost any value of capacitance we need (much like we only have a few kinds of coins and bills in our monetary system, but we can combine them to form any amount). If the capacitors are all connected to the same two points (like across a battery), we’ll call that a parallel arrangement. Take a look at the drawing below for an example:

+

-

Each capacitor is represented by a pair of parallel lines (red in this case) separated by a small gap. This is what an ideal parallel plate capacitor would look like. The parallel lines at the bottom (blue in this drawing) are different in shape – one is long and thin, the other short and thick. This is the symbol for a battery. The other lines are just wires (black in the picture) connecting the circuit elements.

22

PHYS 2212 These capacitors are connected in parallel, but how are they different from a single capacitor with area equal to the sum of their areas? There is no difference. For a given voltage V, the charge on capacitor 1 is Q1 = C1V and the charge on 2 is Q2 = C2V. The total charge is therefore Q1 + Q2 = C1V + C2V = (C1 + C2)V. The capacitance of the two in parallel is then just the sum of their individual capacitances: CTotal = C1 + C2. If the capacitors are in series, interestingly enough, we’ll see that they combine in a very different way. Notice that the right plate of the left capacitor and the left plate of the right capacitor in the drawing below are connected by a wire.

-

+

The plates must have charges which are equal in magnitude but opposite in sign since the arrangement (plates of different capacitors connected by a wire) would be neutral without a battery. That means that the plates facing them also have charges of the same magnitude and opposite sign. We’ll call it Q and label them all

+Q -Q +Q -Q

-

+

Since the voltage is split between the capacitors, the sum of the voltages across each should be the battery’s voltage.

V = V1 + V2 =

Q Q Q + = C1 C2 CTotal

The capacitance of several capacitors in series is found by the math below:

1 CTotal

=

1 1 + + C1 C 2

23

PHYS 2212 Since a battery had to do work to move charges from one plate of a capacitor to the other, energy is stored when charge is separated. The work done to move these charges is not constant, but depends on the potential difference between the plates. Each electron that’s moved changes the potential difference between the plates slightly, and more work must be done to move the next one. We can find the total work in terms of the total charge and the average potential difference, which is just half of the final potential difference (since we started at 0V). The total work done, and therefore the energy stored, is just W = Q * (½ V), but we know that Q = CV, so we can write

1 U = CV 2 2 where U is the energy stored in the capacitor. We could find the energy density in the region between the plates by replacing C with the formula for it in terms of its geometry and dielectric constant, and replacing V with E d. Making those changes to the formula above will still give us total energy, but if we then divide by the volume inside the plates (A d ) we’ll get an energy divided by a volume or an energy density:

u=

U Volume

1  κ ε0 A  2  ( E d) = 2 d 

( Ad)

1 = κ ε0 E2 2

Dielectrics The purpose of a capacitor is to store charge so that it can be released (usually much faster than it was stored) later. For a large value of capacitance, a small potential difference between the plates will allow us to separate a large amount of charge. For this reason, we would like to find a way to increase the capacitance of a capacitor if possible. It turns out that we can do this by filling the space between the two conductors (we’ll use a parallel-plate model again) with a non-conducting material called a dielectric (what would happen if we used a conductor?). When a dielectric is placed between the plates, the electric field will tend to line up the molecules in the dielectric in such a way as to oppose the capacitor’s field. Imagine a molecule (or atom) as a collection of positive and negative charges (protons & neutrons). When placed in an electric field, the electrons will want to move towards the source of the field, since the source is a positive charge, and the protons will want to move in the direction of the field, since field lines end on negative charges. Therefore, the capacitor’s field tends to stretch the atom very slightly. When the positive and negative charges are separated, even if only by a tiny amount, we get a dipole that has its own field. This field will act in a direction opposite to the capacitor’s field, and the overall effect is to produce a smaller total field between the plates. See the figure below for the effect.

24

PHYS 2212

-

+ + + +

-

+ + -

+ + + +

Dipoles (ovals) form in dielectric (green) in response to field of plates. E-field of dipoles (red) opposes field of plates and makes total internal field (blue) smaller Total internal The field between the plates (E) is reduced compared to its value when the space between the plates is empty (E0) by an amount κ, where κ is called the dielectric constant and depends on the composition of the dielectric itself. Our formula for κ is therefore: Strong E-field between plates

κ=

E0 E

Reducing the size of the electric field by placing this material inside the capacitor acts to increase the capacitance of the arrangement. In other words, the same battery can now separate more charge because of the presence of the dielectric. Current When we place a battery in a circuit, it will push electrons from its negative end through the circuit (doing some work, in general) and back to the positive end. This flow of charges is called the current and is measured in coulombs per second, or amperes. For example, your car battery may be able to deliver several hundred amperes of current to your starter. That means that if you picked a point somewhere along the battery cable and could count charges, you’d see hundreds of coulombs of charge go by every second. The formula connecting charge, current, and time is

I=

dq dt

A battery produces direct current (dc), which means the current is always in the same direction (from the positive pole to the negative pole – this is called the conventional current, and the idea dates from a time when people thought positive charges were doing the moving. Today we know that it’s generally the negatively-charged electron which does the traveling, and the flow of electrons is therefore opposite to the direction of the current). The wall outlets provide alternating current (ac) in which the electrons are pushed back and forth, changing their

25

PHYS 2212 direction many times per second. The difficulty or ease with which electrons are set in motion by a potential is described by the resistance of the material containing the electrons. Resistance and Resistivity The resistance of a given object depends on a few factors: first, the composition of the object. Wires are made of copper, not glass. This measurement of the inherent ability of a material to conduct electricity is called resistivity and is usually abbreviated by the letter ρ. For two objects made of the same material but which have different sizes or shapes, resistance will also be different. Resistance is proportional to the length of the wire (or object) and inversely proportional to its area. In other words, a fatter pipe lowers the resistance, and a longer pipe increases it. The formula for resistance can be written as

R=

ρL A

where L is the length of the current path through the object and A is the area of the wire. This tells us that the units of resistivity must be ohms per meter. From the table in your book, you can see that some metals like copper and silver have resistivities lower than 2 x 10-8 Ω*m. At the insulating end of the spectrum, resistivities are in the neighborhood of 1014 Ω*m or higher. This is why wire is typically made of copper (good conductor) covered with plastic (very bad conductor). If the cord for your lamp was not insulated, the two wires would eventually bump into each other and provide a way for the charges to return to the wall without lighting your lamp (short circuit). If you happened to step on both wires, the current would go through your foot! The dependence of resistance on area can be seen when we look at the sizes of wire used in various situations. The wires inside your radio or computer are typically carrying very small currents and therefore can be very small themselves. The main power cable into your house is probably about the size of your wrist since it’s carrying more current. This is why you sometimes hear that extension cords can be dangerous. If you’re using a thin cord (high gauge), you’re putting a resistance in between the power and the device. You can usually feel the cord getting warm. Longer cords should also be fatter to counteract this problem. The next time you’re in a hardware store or home center, ask them for a 50-foot extension cord you can use with your stove or clothes dryer – these don’t exist for this very reason. What causes resistance? It’s collisions between the moving electrons and the rest of the wire (including impurities in the wire). Imagine rolling bowling balls down hills covered with tree stumps. The balls will roll a short distance and then collide with a stump, losing speed and possibly going back uphill briefly. They’ll get downhill eventually, but it will depend on the spacing of the tree stumps. Current in a wire is also sometimes compared to water in a hose. An interesting similarity between the two is that, in both cases, the speed of the particle (water or an electron) is very much less than the speed at which the signal travels. As soon as you flick a light switch, the electricity travels with a speed not far from that of light. The individual electrons have speeds (because of all the collisions slowing them down) that are typically a few centimeters per second or so. The speed difference in a hose is also noticeable – if the hose is 26

PHYS 2212 filled with water already, it will start spraying almost as soon as the faucet is turned on. An individual water molecule, though, takes a while to get from the faucet to the end of the hose. The resistance of most substances is temperature-dependent. At higher temperatures, the resistance increases. For some materials, they can be cooled to a point where all resistance to the flow of current disappears. This is called superconductivity and was discovered in the early 1900’s, but the highest temperatures at which it could be observed were in the neighborhood of 20 K. In the mid 1980’s, new kinds of superconductors were discovered which remained superconducting to around 175 K or so. This is still very cold, but it’s much easier to cool something to 175 K than 20 K! A material that is superconducting at room temperature would be a major breakthrough and would save an unbelievable amount of energy every year which is now lost to heat. Ohm’s Law The flow of charge through a circuit is very much like the flow of water through pipes. The battery is like a pump that pushes water through one end of the pipes and pulls it out the other end. Just as water doesn’t flow through the pipe without friction, electric charges don’t move through ordinary conductors without some loss of energy. This loss is called resistance for the obvious reason that the movement of charges is being resisted by the material they’re moving through. For many materials, the resistance of a given size of material is a constant and we get the famous equation shown below:

V=IR This is known as Ohm’s Law after the person who discovered it. This says that as the potential difference across a piece of material increases, the current through that material increases in direct proportion. If we connect the two ends of a 1.5 V battery to a black box (we don’t know anything about what’s inside – it just has two wires sticking out of it) and get a current of 5 amperes, we can then connect it to a 9 V battery and know that we’ll get a current of 30 amperes if the box obeys Ohm’s law. Not everything obeys Ohm’s law! The units of resistance are volts per ampere (V/A) or ohms. The common symbol for the ohm is the Greek capital omega, or Ω. When we’re selecting wire for a circuit, we generally want it to have the lowest resistance possible (and we’ll typically assume that the resistance of the wire is zero unless we have a particular reason to worry about these tiny resistances). What would our equation above predict for the current produced if we connect the poles of our 9 V battery with a wire of zero resistance? The equation would predict an infinite (or more accurately, undefined) current. What stops this from happening is the fact that the battery has internal resistance of its own, so we don’t have to worry about the math failing. In reality, the internal resistance is usually pretty small, so you can get currents that are very large by “shorting out” the battery. Doing this with a car battery is almost guaranteed to injure someone. Electric Power We can find the power consumed in a circuit ( = the power delivered by a battery) by taking advantage of what we know about energy. We know that moving a charge q through a potential

27

PHYS 2212 difference V requires an energy equal to qV. We also know that energy per time is power. We can combine these to get

P=

∆q V =IV ∆t

The unit of power is still what it was before – 1 watt = 1 J/sec or, in this case, 1 Volt-Ampere. If the device we’re looking at obeys Ohm’s law, we could also write:

V2 P= I R = R 2

Using this, what is the resistance of one of the heating elements in an electric water heater if it consumes 2500 W at 220 V? It’s about 20 Ω. That means the current must be around 11.2 A. EMF If we want to talk about charges moving, rather than sitting still, they need a reason to move. That reason is called the electromotive force or emf. Emf is not really a force, though – it’s just a potential difference between two places (voltage) which, as we know, will cause charges to move in response to it. The symbol for emf is a script capital “E” or E. If we use a battery to provide the emf, it works by means of a chemical reaction that causes electrons to collect at one end and be depleted from the other end. The potential difference between the ends, or emf, is a measure of how badly the electrons “want” to get back together with the ions. If you connect the ends of the battery with a wire so that the electrons have a way to do it, many of them will move along the wire to rejoin ions (don’t do this – it will drain your battery, and for reasons we’ll see later, it will also make the wire dangerously hot). For our purposes, a battery is just a source of emf. Circuits Combining resistors with other electrically useful elements will give us an electrical circuit. The name is significant because there must be a complete path allowing the electrons to travel from a source of emf (a battery, for example) through the various elements and then back to the other end of the emf source. If that path is not complete, we have an open circuit and no work will be done. When we’re designing these circuits, we find that it’s easier to draw a picture of the circuit than it is to list every element by name. Right now, our circuits will be pretty simple. See the table below for pictures

28

PHYS 2212

Element

Symbol

+

Battery

-

Capacitor Resistor Straight Wire (zero resistance) Kirchoff’s Rules For complicated circuits, which may have more than one battery or generator and multiple current paths, we need to use a couple of guiding principles to figure out what’s really going on. These are called Kirchoff’s rules, and the reasoning behind them will be easy to see. First, the junction rule. This says that, for a given junction (place where 2 or more wires connect), the incoming currents must exactly match the outgoing currents. If this were not true, charge would pile up somewhere in the circuit, causing an emf which would oppose the driving emf. This is really just a statement about the conservation of charge, and we can return to our analogy of people in a building. You can be sure that the number of people entering a revolving door is equal to the number of people leaving it. Mathematically, the currents entering a junction will be positive and the currents leaving it will be negative. The sum of these positives and negatives is required to be zero (Rule 1). The loop rule just tells us that energy is conserved. If we follow a charge all the way around a circuit, it will see as many potential drops (across resistors, etc.) as potential jumps (across batteries or generators). We’ll say that the end of a resistor closest to the positive terminal of a battery or generator is itself positive, and the other end is negative. If there is more than one source of emf, we just guess at the direction of current flow and mark our resistors so that it flows from + to -. When a charge (here we’re assuming positive charges to make life simpler) moves from the positive side of a resistor to the negative side, there is a potential drop equal to IR. When it moves through the battery from the negative side to the positive side (after passing through the rest of the circuit), it will experience a jump in potential equal to the battery’s voltage. The sum of potential drops and potential jumps through the circuit will be zero. We have a little freedom with this method, since we can pick the direction of the current if there are multiple sources of emf and we don’t really know which way it is going in the loop. We combine our two rules, applying the junction rule at each junction and the loop rule around every loop. We’ll get multiple equations with multiple unknowns, and we then have to solve them. If, at any point, we get a negative value for one of our currents, it just means we picked the wrong direction for it.

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PHYS 2212 Internal Resistance As mentioned before, a battery (or a generator, for that matter) has some internal resistance of its own. This is usually very small, but because it will effectively be placed in series with the rest of the circuit, there will be a voltage drop across it. The true voltage across the terminals is called the terminal voltage, and will be lower than it would be without the internal resistance. As the resistance of the rest of the circuit decreases, the importance of the internal resistance increases. Higher current through that internal resistance means a higher voltage drop across it. Resistors in Series If we have more than one resistor in a circuit, one way we can combine them is to join them at one end. If we do this, all of the current must pass through each resistor. This obviously acts to increase the resistance of the circuit. There will be a drop in voltage from one side of a resistor to the other. The size of the drop is determined by the current flowing through the resistor and its resistance.

In the diagram above, it is clear that the current goes through each resistor. In this situation, the resistance of the two resistors is equal to the sums of their individual resistances. In other words,

RTotal = R1 + R2 + R3 +  for multiple resistors in series. For example, if the battery produces a potential difference of 18 V and the resistances of the two resistors are 2 Ω and 10 Ω, what will happen in the circuit? The total resistance, by our formula above, must be 2 Ω + 10 Ω = 12 Ω. That means the current must be V/R or 18 V / 12 Ω, or 1.5 A. (Think about what would happen between the resistors if the current were different in each!). For a current of 1.5 A, the voltage drop across the 2 Ω resistor will be 2 Ω times 1.5 A or 3 V. When we get to the next resistor, the voltage drop across it will be 10 Ω times 1.5 A or 15 V. Total voltage drop = 15 V + 3 V = 18 V = battery voltage, so we’re OK. If you’ve seen old Christmas lights, they were built so that one blown light bulb (the filament of which is just a resistor) took out the entire string. That’s because they were wired in series. If you break one part of the circuit, it’s all going to stop. Resistors in Parallel If the resistors in a circuit are arranged so that there are multiple separate paths for the current to take, they are said to be in parallel.

30

PHYS 2212

In the drawing above, a given electron will pass through either one resistor or the other, but not both. This provides a second path for the current to take and therefore reduces the total resistance to its flow. If you imagine people leaving a crowded building, we can think of them as the charges and resistors as the doors. If we have two doors in parallel (meaning you can go through either one to get outside), more people will be able to leave per second than with only one of the doors open. Putting doors in series would correspond to having to use both doors to get out, and that will obviously let fewer people leave than if they only had to go through one door or the other. The voltage drop across each resistor above is the same as the voltage of the battery (they’re both connected to the poles of the battery individually, if we consider the wires to have zero resistance). For different resistances, these identical voltage drops mean different currents pass through each resistor. If we return to our 18 V battery and resistors of 2 Ω and 10 Ω, we’ll find that the current through the 2 Ω resistor is 18 V / 2 Ω or 9 A. The current through the 10 Ω resistor is 18 V / 10 Ω or 1.8 A, giving a total current in the circuit of 10.8 A. If we want to replace this pair with a single resistor which gives the same current flow, it would have to be 18 V / 10.8 A or 1.67 Ω. We can find that directly by learning how to combine resistors in parallel. The equivalent resistance of a number of resistors in parallel is

1 RTotal

=

1 1 1 + + + R1 R2 R3

Let’s check this formula on the result we just found. We should find that 1/RTotal is just the sum of 1 / 2 Ω and 1 / 10 Ω, or 6/10 Ω. RTotal must then be 1/ (6/10 Ω) or 10/6 Ω = 1.67 Ω. (This, by the way, is also the way to determine the mpg a van has to get before it will use less gas on a trip than 2 (or more) cars – remember that for spring break). To summarize, resistors in series all carry the same current, but have voltage drops that depend on their individual resistances. Resistors in parallel are all at the same voltage, but carry a current that depends on their individual resistances. The resistance of resistors in series is always greater than the largest individual resistance. The resistance of resistors in parallel is always smaller than the smallest individual resistance. Now we see why short circuits are dangerous. If the current can go through a wire laid across the battery terminals, the resistance in the wire is very low – close to zero. The circuit has the same resistance it always had. The current now has two choices. It can take a low resistance path, or a high resistance path. It will take the low resistance path and lots of current will flow. We know 31

PHYS 2212 that the power lost to heat in this wire (because it has some resistance, even if it is small) will be V2/R or I2R, which will be very large when R is small. How are the outlets in your house wired? They must be in parallel. If they were in series, all of your lights and appliances would go off when a light bulb blew. This also explains why plugging too many appliances into a circuit will cause your circuit breaker to trip. Each device you add in parallel lowers the total resistance of the circuit and therefore increases the current. When the current reaches a certain value, the circuit breaker will trip and the power will be cut off. This used to be done by fuses (and is still done that way in most cars). The idea behind the fuse is to isolate a small piece of wire made of a material that melts at a low temperature. If that little isolated piece of wire burns through, the circuit is open. It’s much better for that little piece of safely-contained wire to melt than for a random part of the wire inside your wall to do it! Many of the resistive circuits we’ll eventually work with can be reduced to collections of resistors in series and parallel. Whenever you’re trying to decide whether two resistors are in series or parallel, just look at the current path: if the current has to go through both, they’re in series. If it’s required to pick a path, they’re in parallel. Current and Voltage Measurements How can we check ourselves and see if the numbers we calculate for things like currents and voltage drops are accurate? As we’ll see later, a current both responds to and creates its own magnetic field. By arranging a coil of wire between magnets, we can watch it move as a current passes through it. If we put a spring on it to resist its movement and a needle on the coil to track the movement, we’ve made a device called a galvanometer, which is the main part of a current measuring instrument called an ammeter. The other important part is a shunt resistor, which gives the current an alternate path through the device and thereby diverts large currents (most of them, at least) around the sensitive galvanometer. If the galvanometer needle deflects enough to measure small currents, we can see that very large currents would cause the needle to go around many times (or, more likely, would burn out the thin wires inside the galvanometer). As long as we know the relative resistances of the galvanometer and the shunt resistor, we can figure out the total current. The ammeter has to be inside the circuit to measure the current since it can only measure what flows through it. That means it will be in series with the part of the circuit it’s trying to measure. That also means that its presence will increase the total resistance and therefore lower the current. The process of measurement alters the true value we’re trying to measure! This is not really unfamiliar to us, though. If we want to know the temperature of a small cup of coffee and we stick a large, room-temperature thermometer in it, we know from last semester’s study of heat that the coffee will cool a little as it warms the thermometer up. We minimize this effect in the specific heat lab by using quantities of water that are large compared to the mass of the thermometer. We can minimize the effect of our ammeter by making its internal resistance as small as possible. If we connect the galvanometer in parallel with part of the circuit, we will be providing an alternate path for the current. If the galvanometer has a tiny resistance, we’ll siphon off most of 32

PHYS 2212 the current and burn out our ammeter. To get around this, we put a large resistor in series with the galvanometer to make the current path less attractive to the charges in the main circuit. The current that does pass through the galvanometer (deflecting the needle) can be multiplied by the total resistance of the galvanometer + large series resistor to get the voltage between two points. Now we’ve made a voltmeter. Because the addition of a voltmeter gives a new current path, it will alter the behavior of the circuit unless the resistance in the new path is very large. Therefore, we’d like a voltmeter to have a very high resistance, just like we want our ammeter to have very low resistance. Remember that ammeters are wired in series with the circuit, and voltmeters are wired in parallel with it. The ammeter can be thought of as replacing a tiny segment of the wire in the circuit. The voltmeter is connected between two different points to measure the potential difference. If we do this backwards, we’ll have a problem. A voltmeter in series with a circuit will make the total resistance huge and will probably stop the current. An ammeter in parallel with a circuit will (because of its low resistance) suddenly find itself routing a huge current between two points. For this reason, most ammeters have fuses that are designed to burn out first before the galvanometer can be destroyed. RC Circuits We’ve glossed over this before, but when we connect a capacitor to a circuit, the charge that it separates doesn’t collect instantly. As charges flow from one end to the other through the battery, they collect on the plates and begin to establish an electric field that opposes the accumulation of more charges. This shouldn’t be surprising – we know that the negative charges on one plate would like to be as far from each other as possible. The charge on the capacitor builds up to its theoretical maximum value of Q0 = C V as time passes according to the formula below:

(

Q( t ) = Q0 1 − e −t / RC

)

In other words, the rate at which the capacitor charges depends on the resistance and capacitance of the circuit. How long would it take to fully charge? According to the formula (and assuming charge doesn’t come in fixed-size units of electrons), an infinite amount of time. As a practical matter, we’ll see that Q is close enough to Q0 after a few time constants have passed. The time constant for the circuit is found by just multiplying resistance and capacitance:

τ = RC After one time constant (so t = τ ), the value of the charge will be Q0 (1 – e-1) or about 0.63 Q0. When t = 5 τ, we’ll get Q0 (1 – e-5) or about 0.993 Q0. The capacitor will discharge exponentially as well. Assuming it starts off charged to Q0, it will discharge according to

33

PHYS 2212

Q( t ) = Q0 e −t / RC The combination of resistor and capacitor makes a useful device for measuring time electronically. Since the charge held by the capacitor is changing with time, so will the voltage between the plates as well as the current used to deliver charge to the plates. These can be easily found for either the charging or discharging cases by the use of the formulas we’ve already seen:

V=

q C

and

34

i=

dq dt