Spherical conducting shell enclosing a non-conducting core

  • Thread starter Thread starter hmparticle9
  • Start date Start date
  • Tags Tags
    Electrostatic
AI Thread Summary
In the discussion about a spherical conducting shell enclosing a non-conducting core, it is established that the charge inside the outer shell arranges itself to neutralize the enclosed charge as much as possible. For the region between the inner boundary and outer shell, the electric field is zero due to the properties of conductors. When considering a scenario with less charge than expected on the outer shell, the electric field can be calculated using Gauss's law, which indicates that the electric field depends on the difference between the total charge and the charge accumulated at the inner boundary. The direction of the electric field within the conductor is clarified, emphasizing that it must be zero because conductors are equipotential. This leads to a deeper understanding of how electric fields behave in such configurations.
hmparticle9
Messages
151
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?
 
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.
 
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
TL;DR Summary: Find Electric field due to charges between 2 parallel infinite planes using Gauss law at any point Here's the diagram. We have a uniform p (rho) density of charges between 2 infinite planes in the cartesian coordinates system. I used a cube of thickness a that spans from z=-a/2 to z=a/2 as a Gaussian surface, each side of the cube has area A. I know that the field depends only on z since there is translational invariance in x and y directions because the planes are...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Back
Top