(General question)Area charge density and electric potential

[prodo123]

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]

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 ?

 

[prodo123]

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!