How To Find the Electric Potential for a Given Charge
Distribution
Here are two methods; sometimes both work, and sometimes one is easier,
depending on what you are given.
Method 1: Superposition Method
This method is very similar to the method used for finding the electric
force or electric field from a set of charges by superposition,R as given in the
continuous distribution of charge howto, but with V = k dqr . The main
difference is that potential is a scalar, not a vector, so you need to add up
scalar rather than vector contributions. For a positive charge, potential is
positive, and for a negative charge, potential is negative. Use this method
for point charges, or for a continuous distribution for which you know the
contributions to the potential from pieces of it that you can add up.
Method 2: Electric Field Integration Method
Use this method if you have a situation for which you know the electric
field (you are told it or can find it easily) and are asked for a potential or a
potential difference. It tends to work well for cases of spherical or cylindrical
symmetry (or those for which Gauss’s Law works well.)
1. First, find the electric field in the region of interest (often, you can use
Gauss’s Law...follow the Gauss’s Law howto.)
2. Choose a point such that V = 0. You’re free to define this to be any
point in space. Often, it’s convenient to choose V = 0 at infinity.
3. Now, set up a line integral which gives the potential difference ∆V
between two points A and B.
∆V = VB − VA = −
Z B
~ · d~l,
E
(1)
A
This involves choosing:
• Point B, where you want to find the potential.
• Point A, a point where you know the potential already. This could
be a point where you have defined the potential to be zero (often
it’s convenient to pick this point to be ∞), or it could be a point
where you have already calculated the potential.
• A path to integrate along. To do the integral easily, choose a path
~ · d~l just becomes Edl.
that’s parallel to the electric field. Then E
What you are doing here is adding up all the potential differences along
the line.
4. Plug in the electric field over the region where your line integral path
runs, and do the integral.
5. Solve for VB .