# Electric Potential of a Uniformly Charged Sphere

1. Mar 25, 2013

### Typhon4ever

My book gives the electric potential of a uniformly charged sphere as $\frac{1}{4\pi \epsilon_0}$$\frac{Qr^2}{r_0^3}$. I can't derive this forumula. What I get is $\frac{1}{8\pi \epsilon_0}$$\frac{Qr^2}{r_0^3}$. I can guess how the book did it which is by taking the equation for voltage of a single point charge $V=\frac{1}{4\pi \epsilon_0}$$\frac{Q}{r}$ then subbing in for Q enclosed $Q\frac{r^3}{r_o^3}$ which is the charge in terms of r. That would give you the book answer. I tried doing it a different way however which I'm not sure why it won't work. I used Va-Vb=$\int _a^{b}$ E dl. Yet when I get to solving for the electric field I get $\frac{Q}{4\pi r^2 \epsilon_0}$. If I take Q out then I can solve for the book answer. However, Q is in terms of r which I wrote above so I can't take Q out. I must integrate it. That gives me the 1/8 instead of the 1/4. What am I doing wrong? Thanks!

2. Mar 26, 2013

### Staff: Mentor

Do you have a uniformly charged sphere, or ball? The derivation suggests a ball (charge everywhere in the volume), not a sphere. Q is the total charge, and r_0 is the radius of the ball?
How do you get the electric field? It should not depend on r like that.

3. Mar 29, 2013

### Chandra Prayaga

Are you looking for the potential inside the sphere or outside? In other words is r > r0 or is r < r0?