Spherical shells (inner conducting and outer nonconducting)

[goohu]

Homework Statement

See picture

Relevant Equations

V = $-\int E dl$

a) I think you find V by just integrating E in regards to R. Then we integrate from the point of interest, which is a, to the 0 potential which is (R = 2a)?

b) If the same logic as a) applies here as well then we should integrate from the point of interest to the 0 potential. This should be from 0 to 2a.

 

[haruspex]

I think you find V by just integrating E in regards to R

Yes, except that R is a given constant, so better to say wrt some new variable, r say.

b) If the same logic as a) applies here as well then we should integrate from the point of interest to the 0 potential.

Yes, except that there is an easier way. Hint: it's a conductor.

 

[goohu]

R is a constant? Can you explain why? E(R) seems to be a function depending on R, for me it seems like its a variable? I don't understand. What do you mean with wrt?

For b) if its a conductor you could use Gauss law I guess. C = Q/V

I'm mostly familiar with using it to determine the capacitance between two conducting plates. Usually you find E(R) then you integrate to find V and integrate from 0 to d , if d is the distance between the plates.

In this case it's from one conducting shell to origin. I'm unsure about the integration limits.
Is it from 0 to a, or a to 0?

 

[haruspex]

R is a constant?

Sorry, dealing with too many threads at once.

For b, what do you know about the potential within a conductor?

 

[goohu]

Sorry for the late reply.

b)
The electric field inside a conductor (in this case 0 to a) is 0. So V should be a constant. Not sure if we need to go further than this?

I am unsure because of the wording in the task : "using the same assumptions as those of part (a)."
The assumption being "the electrostatic potential on the outer surface of the non-conducting material is 0 V." The stated assumption shouldn't affect the problem?

 

[TSny]

b)
The electric field inside a conductor (in this case 0 to a) is 0. So$V$ should be a constant. Not sure if we need to go further than this?

Yes, $V$ is constant throughout the conductor. You need to express V at the center of the conductor in terms of $\varepsilon_0, a$ and $\rho$.

I am unsure because of the wording in the task : "using the same assumptions as those of part (a)."
The assumption being "the electrostatic potential on the outer surface of the non-conducting material is 0 V." The stated assumption shouldn't affect the problem?

The choice of where $V = 0$ does affect the answer to part (b). For part (b) you want to still take $V=0$ at the outer surface of the non-conductor.

 

[goohu]

There is the equation for V. I assume both shells has the same charge density (a bit unclear to me).

$V = \int \frac{\rho_v}{4\pi\epsilon_0R} dV = \int \frac{\rho_v}{4\pi\epsilon_0R} R^2 sin\theta dRd\phi d\theta = \frac{\rho_v}{\epsilon_0} \int R dR = \frac{\rho_v a^2}{2\epsilon_0}$ (integrating from 0 to a)

I am unsure where to go from here?

 

[TSny]

a) I think you find V by just integrating E in regards to R. Then we integrate from the point of interest, which is a, to the 0 potential which is (R = 2a)?

Yes. What do you get for an answer?

b) If the same logic as a) applies here as well then we should integrate from the point of interest to the 0 potential. This should be from 0 to 2a.

As @haruspex stated, you could do this but there is an easier way. As you stated in post #5, you know that V is constant throughout the conductor. So, how does the potential at the center of the conductor compare to the potential at the surface of the conductor?

 

[goohu]

a) we are integrating from a to 2a.
$V = -\int E dl = -\int \frac{R^3 \rho}{3\epsilon_0 R^2} + \int \frac{a^3 \rho}{3\epsilon_0 R^2} = - \frac {\rho}{3 \epsilon_0} \int R dR + \frac{a^3 \rho}{3 \epsilon_0} \int \frac{1}{R^2} dR = -\frac{\rho a^2}{2\epsilon_0} + \frac{a^2 \rho}{6\epsilon_o}$

b) Let's assume we integrate from 0 to 2a. Shouldnt the answer be the same as a) if the electric field is 0 inside the conductor? This already seems easy enough to me.

But to answer your question, I think it should be the same?

 

[TSny]

a) we are integrating from a to 2a.
$V = -\int E dl = -\int \frac{R^3 \rho}{3\epsilon_0 R^2} + \int \frac{a^3 \rho}{3\epsilon_0 R^2} = - \frac {\rho}{3 \epsilon_0} \int R dR + \frac{a^3 \rho}{3 \epsilon_0} \int \frac{1}{R^2} dR = -\frac{\rho a^2}{2\epsilon_0} + \frac{a^2 \rho}{6\epsilon_o}$

b) Let's assume we integrate from 0 to 2a. Shouldnt the answer be the same as a) if the electric field is 0 inside the conductor? This already seems easy enough to me.

But to answer your question, I think it should be the same?

For part (a) what limits of integration did you use?

Part (b) is really that easy.

 

[goohu]

for a) integration limits were a to 2a.

 

[TSny]

for a) integration limits were a to 2a.

Getting the signs correct can be tricky. Recall that the basic relation is

$V(B) - V(A) = -\int_A^B \mathbf E \cdot d\mathbf l$

If you integrate from $a$ to $2a$, what does the left-hand side of the above relation become? In particular, which of $V(A)$ and $V(B)$ would be zero and which would be the potential at the point that you are interested in?