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)

2.5. Apparent Answer

(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.

 

[SammyS]

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)

2.5. Apparent Answer

(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.

 

[ehild]

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.

 

[ehild]

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