- #1

prodo123

- 17

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## Homework Statement

Sorry, the post isn't about a single homework problem but rather something that I keep getting confused on. It's about calculating the electric potential of a spherical shell of uniform charge in two different ways.

## Homework Equations

##\Delta V=\int_a^b -\vec E\cdot d\vec r##

##V=\frac{1}{4\pi\epsilon_0}\int\frac{dq}{r}##

Gauss's Law

## The Attempt at a Solution

Taking ##V=0## at ##r\to\infty##, the electric potential of a spherical shell of charge ##Q## with radius ##R## should be:

##V=\int^R_\infty -\vec E\cdot d\vec r##

Using Gauss's Law, the electric field outside the shell is ##\vec E = \frac{Q}{4\pi \epsilon_0 r^2}\hat r##. This gives ##V=\frac{Q}{4\pi\epsilon_0}(\frac{1}{R}-\frac{1}{\infty})=\frac{Q}{4\pi\epsilon_0 R}##, which is what we expect.

The charge area density for the shell is ##\sigma=\frac{Q}{4\pi R^2}=\frac{q}{A}##. Having A be the surface area for a sphere, the function for the charge of a spherical shell of radius r is:

##q(r)=\sigma A = \frac{Q}{4\pi R^2}(4\pi r^2)=Q\frac{r^2}{R^2}##

The infinitesimal charge is therefore ##dq=\frac{2Q}{R^2}r dr##.

Since electric potentials for a distribution of charge is found by ##V=\frac{1}{4\pi\epsilon_0}\int\frac{dq}{r}##, the ##r## variables cancel out and the integral reduces to ##V=\frac{1}{4\pi\epsilon_0}\int^R_0\frac{2Q}{R^2}dr##.

Evaluating the integral gives ##V=\frac{Q}{2\pi\epsilon_0 R}##, which is not what's expected. There's a factor of 2 somewhere that's throwing off the answers. Where am I going wrong?