# (General question)Area charge density and electric potential

## 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?

BvU
Homework Helper
a spherical shell of uniform charge
How thick is the shell ? You integrate from 0 to R but is that really where the charge resides ?

How thick is the shell ? You integrate from 0 to R but is that really where the charge resides ?
##A_{\text{sphere}}=R^2\iint sin(\theta)d\theta d\phi## where ##\pi \ge \theta \ge 0## and ##2\pi\ge\phi\ge 0## for an infinitely thin spherical shell
Leaving ##\phi## as a variable gives the following:
##A=2R^2\int d\phi##
##dA = 2R^2 d\phi##
##q=\sigma A##
##dq = \sigma dA = 2\sigma R^2 d\phi##
##V=\frac{1}{4\pi\epsilon_0}\int^{2\pi} _0 (\frac{1}{R})2\sigma R^2 d\phi##
##V=\frac{\sigma R}{\epsilon_0} = \frac{\frac{Q}{4\pi R^2}R}{\epsilon_0}##
##V=\frac{Q}{4\pi\epsilon_0 R}##
Right...thanks!