Worksheet 4: The Electric Potential. Work and Energy. First let's investigate how
the electric potential energy changes. Remember that. WF = ∫ b. aF · ds = -∆UF.
Worksheet 4: The Electric Potential Work and Energy First let’s investigate how the electric potential energy changes. Remember that
�
WF =
b a
F · ds = −∆UF
1. Imagine a proton and an electron, a distance r apart. What force does the proton’s electric field exert upon the electron? Don’t use numeric values; e.g. use ±e for the charges. This is Coulomb’s Law; F = ke q1 q2 /r 2 , so: 2
� = −ke e rˆ F r2
2. Let’s pretend that the proton is fixed in space, but the electron is free to move. The electron will be drawn directly towards the proton. If it moves from a distance of R1 to R2 , how much work is done upon it? To get the right sign we have to be careful with the integral. The force will act directly along the path, and in the same direction as the displacement. So the integral for work will be � �
rˆ � · ds r2
� = −ke e2 � · ds F
� will point towards the proton; so does rˆ. ds −ke e2
�
R2 R1
dr = ke e 2 r2
�
1 1 − R2 R1
�
3. What is the change in potential energy of the system? From above, just opposite to the work done.
∆U = ke e2
�
1 1 − R1 R2
�
4. If the electron is brought from very, very far away to a distance R from the proton, what would be the change in potential energy? In the limit when R1 gets very large, 1/R1 → 0, so
∆U = −ke e2
1
1 R2
The Electric Potential Let’s say that the potential has a value of V = 0 infinitely far away from the proton. And let’s forget about the electron! We don’t need it to talk about the potential. 5. What is the value of the electric potential a distance R away from the proton? In this worksheet we were only using the definition of potential in terms of potential energy. So the potential from the proton is just U/q , where q = −e (The charge of the electron). So
V = ke
e R
6. What is the value of the electric field at such a point? Is there an obvious connection between them? We already know that the electric field is
� = ke e rˆ E R2 It turns out that Er = − ∂E ; the electric field is related to the derivative of the potential. ∂r
