Spherical shells (inner conducting and outer nonconducting)

In summary, the conversation discussed the integration of electrostatic potential in regards to a given constant R. It was concluded that for a non-conducting material, the potential at the outer surface is 0 V and for a conducting material, the potential is constant throughout. In part a), the integration was done from a to 2a, while in part b), integrating from 0 to 2a would yield the same result. However, there is an easier method to solve part b) by considering the constant potential throughout the conductor.
  • #1
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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.
 

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  • #2
goohu said:
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.
goohu said:
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.
 
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  • #3
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?
 
  • #4
goohu said:
R is a constant?
Sorry, dealing with too many threads at once.

For b, what do you know about the potential within a conductor?
 
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  • #5
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?
 
  • #6
goohu said:
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.
 
  • #7
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?
 
  • #8
goohu said:
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?
 
  • #9
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?
 
  • #10
goohu said:
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. :smile:
 
  • #11
for a) integration limits were a to 2a.
 
  • #12
goohu said:
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?
 

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