Potential of Concentric Spherical Insulator and Conductor

In summary, the problem involves a solid insulating sphere with a uniform charge density, surrounded by a conducting shell of different radii. The x-component of the electric field at a point located a certain distance away is requested, as well as the electric potential at various points on the sphere and shell. When attempting to solve for the potential at the inner and outer surfaces, the instructions suggest that the conducting shell plays a role and to revisit the definition of potential.
  • #1
ml8
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Homework Statement



A solid insulating sphere of radius a = 5.6 cm is fixed at the origin of a co-ordinate system as shown. The sphere is uniformly charged with a charge density ρ = -494 μC/m3. Concentric with the sphere is an uncharged spherical conducting shell of inner radius b = 10.8 cm, and outer radius c = 12.8 cm.

1) What is Ex(P), the x-component of the electric field at point P, located a distance d = 31 cm from the origin along the x-axis as shown?
2)
What is V(b), the electric potential at the inner surface of the conducting shell? Define the potential to be zero at infinity.
3)
What is V(a), the electric potential at the outer surface of the insulating sphere? Define the potential to be zero at infinity.
4)
What is V(c) - V(a), the potentital differnece between the outer surface of the conductor and the outer surface of the insulator?
5)
A charge Q = 0.0383μC is now added to the conducting shell. What is V(a), the electric potential at the outer surface of the insulating sphere, now? Define the potential to be zero at infinity.

Homework Equations



|[itex]\vec{E}[/itex]|=kQ[itex]_{enc.}[/itex]/r[itex]^{2}[/itex]
V=∫kQ[itex]_{enc.}[/itex]/x[itex]^{2}[/itex]dx

The Attempt at a Solution



1)E[itex]_{x}[/itex](P)= -33995.1 N/C
Q=[itex]\rho[/itex]V=-3.634E-7C
E=kQ/r[itex]^{2}[/itex]

When I try to complete 2) I am given the message: It looks like you have calculated the potential at the inner radius of the shell to be equal to the potential at r = c produced by the insulating sphere by itself. The conducting shell plays a role here. Go back to the definition of the potential to determine the answer.
When I try to complete 3) I am given the message: It looks like you have calculated the potential at the outer radius of the insulating sphere to be equal to the potential at r = a produced by the insulating sphere by itself. The conducting shell plays a role here. Go back to the definition of the potential to determine the answer.
 
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  • #2
Hello ml8,

Welcome to Physics Forums! :smile:

ml8 said:
When I try to complete 2) I am given the message: It looks like you have calculated the potential at the inner radius of the shell to be equal to the potential at r = c produced by the insulating sphere by itself. The conducting shell plays a role here. Go back to the definition of the potential to determine the answer.
Please show us what you've done with 2).

When I try to complete 3) I am given the message: It looks like you have calculated the potential at the outer radius of the insulating sphere to be equal to the potential at r = a produced by the insulating sphere by itself. The conducting shell plays a role here. Go back to the definition of the potential to determine the answer.
Please show us what you've done with 3).
 

FAQ: Potential of Concentric Spherical Insulator and Conductor

1. What is the purpose of using a concentric spherical insulator and conductor?

The purpose of using a concentric spherical insulator and conductor is to create a controlled electric field within a specific area. This is important for a variety of applications, such as in electrical insulation, capacitors, and high voltage transmission lines.

2. How does a concentric spherical insulator and conductor work?

A concentric spherical insulator and conductor work by utilizing the principle of equipotentiality. This means that all points on a conductor have the same potential, and by surrounding it with an insulator, the electric field is confined to the space between the two spheres, providing a controlled environment for electric charges.

3. What materials are commonly used for concentric spherical insulators and conductors?

The most commonly used materials for concentric spherical insulators and conductors are metals for the conductor, such as copper or aluminum, and non-metallic materials for the insulator, such as glass, porcelain, or rubber. The specific materials chosen depend on the application and required properties.

4. Can a concentric spherical insulator and conductor be used for both AC and DC applications?

Yes, a concentric spherical insulator and conductor can be used for both AC (alternating current) and DC (direct current) applications. The design and properties of the insulator and conductor must be carefully chosen to ensure proper functioning and safety for the specific type of current being used.

5. What are the advantages of using a concentric spherical insulator and conductor?

Using a concentric spherical insulator and conductor provides several advantages, including better control over electric fields, increased safety and reliability, and the ability to withstand high voltages. It also allows for more compact designs and efficient use of materials.

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