# Calculating Potential from Field

1. Feb 14, 2010

### Ron Burgundy

1. The problem statement, all variables and given/known data

A nonconducting sphere has radius R=2.31 cm and uniformly distributed charge q=+3.50 fC. Take the electric potential at the sphere's center to be V0=0. What is V at radial distance r = 1.45cm?

2. Relevant equations
E= $$\frac{kqr}{R^3}$$

$$\Delta$$V=-$$\int$$E*ds

3. The attempt at a solution

E = $$\frac{k(3.5*10-15C)(.0145m)}{(.0231m)^3}$$
E = 3.70*10-2 V/m

$$\Delta$$V=-$$\int$$3.70*10-2 V/m * ds
$$\Delta$$V=-3.70*10-2 V/m * .0145 m

$$\Delta$$V = -5.365*10-4

That's not right according to the book. I keep getting double the right answer when I do problems with electric potential and I can't figure out why.

2. Feb 14, 2010

### kuruman

The problem asks you to find V not ΔV. Don't forget that the electric potential is zero at infinity. To do it right you must take the integral from infinity to 1.45 cm which means that you need to do two integrals because the electric field changes at r = 2.31 cm.

3. Feb 14, 2010

### Ron Burgundy

That doesn't really make sense to me. Are you saying I should take the integral from infinity to .0231 m using $$\frac{kq}{r^2}$$ for the electric field and then add the integral from .0231 m to .0145 m using $$\frac{kqr}{R^3}$$? That doesn't produce the right answer. However, using the integral from 0 to .0145 m of $$\frac{kqr}{R^3}$$ works. The problem is that I'm still not really understanding why it works.

4. Feb 14, 2010

### ideasrule

No, I think kuruman misread the question. The question does ask for delta-V, not for V.

Electric field is kq/r^2, but q is not constant; the larger r is, the larger q is. Try expressing q in terms of R, r, and Q.

5. Feb 14, 2010

### kuruman

Sorry, I misread the question. I agree with ideasrule. The electric field inside the sphere is not constant. If E = kQr/R3 inside, then that is the expression that you must put in the integral and integrate from zero to 2.31.