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Electric potential inside a shell of charge

  • Thread starter prodo123
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  • #1
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Homework Statement


Q1: There are two concentric spherical shells with radii ##R_1## and ##R_2## and charges ##q_1## and ##q_2## uniformly distributed across their surfaces. What is the electric potential at the center of the shells?

Q2: There is an infinitely long hollow cylinder of linear charge density ##\lambda## and radius ##R##. What is the potential difference ##\Delta V## between the surface of the shell and a radius ##R'## inside the cylinder?

Homework Equations


##\vec E = -\nabla V##
##\oint \vec E \cdot d\vec A = \frac{Q\text{encl}}{\epsilon_0}## (Gauss's Law)
##V=\frac{q}{4\pi \epsilon_0 r}##

The Attempt at a Solution



Starting with Q2, Gauss's Law using a cylinder as the Gaussian surface shows there is no enclosed charge; ##\vec E = \vec 0##. Because ##\vec E = -\nabla V##, one can conclude that ##V=0## between ##R'## to ##R##. This is the given (and found) answer for Q2.



In a similar way, there is no enclosable charge for all points inside the two spherical shells in Q1. By the same logic as Q2, it would seem ##\vec E=0=-\nabla V## and there would be no electric potential at the center of the shells.

However, Gauss's Law cannot be applied to a point or line since the Gaussian surface has area ##A=0## and Gauss's Law reduces to ##0=0##. Therefore, one cannot find the electric field at the center of the sphere, and ##\vec E = -\nabla V## cannot be used.

Since the center of the shells are at a constant distance ##R_1## and ##R_2##, the electric potential can be found by:

##V=\frac{q_1}{4\pi \epsilon_0 R_1} + \frac{q_2}{4\pi \epsilon_0 R_2}##

which is the given answer for Q1. (One needs to integrate the charge density across a spherical area, which ultimately reduces to the answer above)



This result (i.e. textbook answers) seems to show some weird results:
  • The electric field and potential are zero for all positions inside a closed area of charge and nonzero at the symmetrical center or axis.
  • The graphs of the field magnitude and electric potential are discontinuous at the center.
  • If this is true, the electric field vector there has no defined direction...? (or is undefined since the equation is discontinuous)
  • In Q2, the electric field and potential should be nonzero along the axis of the cylinder and zero for all spaces between the axis and the cylinder wall.

I'm somewhat confused because the two questions seem to contradict each other. Is my logic correct in interpreting the answers?
 

Answers and Replies

  • #2
Doc Al
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One note: Between points where the field is zero there is no change in potential. That does not mean that the potential is zero. (There can certainly be a potential with respect to some other point.)
 
  • #3
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One note: Between points where the field is zero there is no change in potential. That does not mean that the potential is zero. (There can certainly be a potential with respect to some other point.)
Integrating ##\vec E \cdot d\vec r## where ##\vec E = \vec 0## to find the potential at a single point results in ##V = 0+C##. Then the nonzero potential found at the center is ##C## and is constant across the space inside the shells...

Makes much more sense, thanks!
 
  • #4
rude man
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Or perhaps:In Q2: the E field just outside the surface is σ/ε where σ is surface charge density (related to λ obviously). The E field just below the surface is zero. Both by Gauss. Since potential is the integral of the E field over distance, and the distance → zero, therefore there is no change in potential between the outside & inside surfaces.

Q1: Can also do this by superposition theorem:
Potential of shell 1 with q2=0 is kq1/R1.
Potential of shell 2 with q1=0 is kq2/R2.
Total potential is sum of above potentials.
 

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