# Relating Electric Field and Voltage in an Insulating Sphere

NeonJay

## Homework Statement

Positive electric charge Q is distributed uniformly throughout the volume of an insulating sphere with radius R. From the expression for E=kQ/r^2 for r>R and E=kQr/R^3 for r<R, find the expression for the electric potential V as a function of r both inside and outside the uniformly charged sphere. Assume that V = 0 at infinity.

k=1/(4*π*ε0)

## Homework Equations

$$\int_A^B\vec{E}\cdot\mathrm{d}\vec{r} = V(A) - V(B)$$

(kQ/2R)[3-(r^2)/(R^2)]

Apologies ... I'm not very knowledgeable in tex, but that last part would read r-squared over R-squared.

## The Attempt at a Solution

I have already obtained V = kQ/r for r>R. This was a simple antiderivative.
I'm getting really caught up on r<R. I've tried integrating from 0 to r, 0 to R, r to R, and backwards versions of the same. Nothing comes to the answer in 2.5 above. I'm really at a loss for how this integral should be done. I can understand most of it except for the 3 in the second expression; where the heck does it come from? I've seen this answer in a couple of different places, but none of them actually explain it.

Any help would be appreciated. Thanks ahead of time.

Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

Positive electric charge Q is distributed uniformly throughout the volume of an insulating sphere with radius R. From the expression for E=kQ/r^2 for r>R and E=kQr/R^3 for r<R, find the expression for the electric potential V as a function of r both inside and outside the uniformly charged sphere. Assume that V = 0 at infinity.

k=1/(4*π*ε0)

## Homework Equations

$$\int_A^B\vec{E}\cdot\mathrm{d}\vec{r} = V(A) - V(B)$$

(kQ/2R)[3-(r^2)/(R^2)]

Apologies ... I'm not very knowledgeable in tex, but that last part would read r-squared over R-squared.

## The Attempt at a Solution

I have already obtained V = kQ/r for r>R. This was a simple antiderivative.
Hello NeonJay. Welcome to PF !

Extend this to r = R, to get V(R) = kQ/R.

Then use $\displaystyle \int_A^B\vec{E}\cdot\mathrm{d}\vec{r} = V(A) - V(B)$ to get V(r) for r < R . To do this, A = R, B = r .
I'm getting really caught up on r<R. I've tried integrating from 0 to r, 0 to R, r to R, and backwards versions of the same. Nothing comes to the answer in 2.5 above. I'm really at a loss for how this integral should be done. I can understand most of it except for the 3 in the second expression; where the heck does it come from? I've seen this answer in a couple of different places, but none of them actually explain it.

Any help would be appreciated. Thanks ahead of time.

Homework Helper
The potential is continuous function, and you know it at r=R. V(R)=kQ/R2 . Integrate from r to R:
$$V(r)-V(R)=\int_r^R{\frac{kQ}{R^3}rdr}$$
substitute kQ/R for V(R) and solve for V(r).

ehild

NeonJay
I think I get it. I was essentially neglecting part of my equation.

Thanks for the help! It really clears a lot of things up.

Homework Helper
The equation comes from the definition of the electric potential: a function V(r) of position r so as the negative gradient of V is equal to the electric field intensity.

-V(r)=E.

The electric field is conservative so the work done on unit charge between two points A and B does not depend on the path taken.

$$\int_A^B{-\nabla V d\vec r}=\int_A^B{\vec Ed\vec r}$$

V dr=dV, the increment of V.

$$\int_A^B{-dV}=V(A)-V(B)=\int_A^B{\vec Ed \vec r}$$

ehild