Potential of Concentric Spherical Insulator and Conductor

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. 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.

http://ift.tt/1hR3Qap

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