Calculating Potential from Field

[Ron Burgundy]

Homework Statement

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?

Homework Equations

E= $$\frac{kqr}{R^3}$$

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

The Attempt at a Solution

E = [tex]\frac{k(3.5*10^-15C)(.0145m)}{(.0231m)^3}[/tex]
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.

 

[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.

 

[Ron Burgundy]

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.

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.

 

[ideasrule]

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 $$\frac{kqr}{R^3}$$?

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

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.

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.

 

[kuruman]

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

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.