Spherical conducting shell enclosing a non-conducting core

  • Thread starter Thread starter hmparticle9
  • Start date Start date
  • Tags Tags
    Electrostatic
Click For Summary
SUMMARY

The discussion focuses on the behavior of electric fields within a spherical conducting shell that encloses a non-conducting core with charge -Q. It is established that for the region r ∈ [b,c), the electric field E(r) is zero due to the charge distribution on the conducting shell, which arranges itself to cancel the internal charge. The remaining charge Q resides on the outer shell at r = c, leading to the expression E(c) = Q/(4 π ε₀ c²). The application of Gauss's law is critical in determining the electric field within the conducting material and at the boundaries.

PREREQUISITES
  • Understanding of Gauss's law in electrostatics
  • Knowledge of electric fields and equipotential surfaces
  • Familiarity with spherical coordinates in electrostatics
  • Basic concepts of charge distribution in conductors
NEXT STEPS
  • Study the implications of Gauss's law for different charge configurations
  • Explore the properties of electric fields in conductors under static conditions
  • Investigate the behavior of electric fields in non-conducting materials
  • Learn about the concept of equipotential surfaces and their significance in electrostatics
USEFUL FOR

This discussion is beneficial for physics students, electrical engineers, and anyone studying electrostatics, particularly those interested in the behavior of electric fields in conductive materials.

hmparticle9
Messages
157
Reaction score
26
Homework Statement
A solid non-conducting sphere of uniform charge density and total charge ##-Q## and radius ##a## is surrounded by a concentric conducting spherical shell of inner radius ##b## and outer radius ##c## with ##a < b < c##. The outer shell has charge##2Q##. I am interested in the case ##r \in [b,c]##
Relevant Equations
$$\int_S \mathbf{E} \cdot d \mathbf{S} = \frac{q}{\epsilon_0}$$
The case I am interested in is ##r \in [b,c]##. Because the outer shell is conducting and the outer shell encloses a charge ##-Q## would it be correct to say that for the case ##r \in [b,c)## the charge inside the outer shell arranges itself in a way to "cancel" as much of the contained charge as it can.

Because the outer shell has ##2Q## to play with we are okay. ##Q## of the ##2Q## charge arranges itself on the shell boundary ##r = b##, hence ##E(r) = 0, r \in [b,c)##.

Am I correct in saying that the remaining charge ##Q## arranges on the outer shell ##r=c##? When we use Gauss's law we obtain
$$E(c) = \frac{Q}{4 \pi \epsilon_0 c^2}$$

If we had less than ##Q## charge in the outer shell I am trying to figure out what ##E(r)## would look like in the outer shell...
 
Last edited:
Physics news on Phys.org
Start by considering what the electric field is at ##b<r<c##, i.e. inside the conducting material. What does Gauss's law have to say about that?
 
  • Like
Likes   Reactions: DaveE
Surely the charge ##\hat{Q} < Q## would accumulate at ##r = b##. Gauss's law would say

$$\int_S \mathbf{E} \cdot d \mathbf{S} = \frac{Q - \hat{Q}}{\epsilon_0}$$ within the shell?
 
Last edited:
hmparticle9 said:
Surely the charge ##\hat{Q} < Q## would accumulate at ##r = b##. Gauss's law would say

$$\int_S \mathbf{E} \cdot d \mathbf{S} = \frac{Q - \hat{Q}}{\epsilon_0}$$ within the shell?
Sure, but you didn't answer my question. What is the value of ##\mathbf E## on the left-hand side of the equation? Hint: Under static conditions, a conductor is an equipotential.
 
Sorry ! :)

$$\int_S \mathbf{E} \cdot \text{d} \mathbf{S} = E(r) 4 \pi r^2 = \frac{Q - \hat{Q}}{\epsilon_0} \implies E(r) = \frac{Q - \hat{Q}}{4\pi r^2\epsilon_0}$$
 
hmparticle9 said:
Sorry ! :)

$$\int_S \mathbf{E} \cdot \text{d} \mathbf{S} = E(r) 4 \pi r^2 = \frac{Q - \hat{Q}}{\epsilon_0} \implies E(r) = \frac{Q - \hat{Q}}{4\pi r^2\epsilon_0}$$
That is not correct.

What is the direction of the electric field inside the conductor? Remember that electric field lines point from regions of high electric potential to regions of low electric potential and that conductors are equipotentials. Put it together and come up with a numerical value for the normal component of the electric field on the surface of the Gaussian surface inside the conductor.
 

Similar threads

Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 23 ·
Replies
23
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
12
Views
2K
Replies
4
Views
2K
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K