Potential of Concentric Spherical Insulator and Conductor

AI Thread Summary
The discussion centers on calculating the electric field and electric potential for a solid insulating sphere and a surrounding conducting shell. The x-component of the electric field at a point 31 cm from the origin is determined to be -33995.1 N/C, and the charge of the insulating sphere is calculated to be -3.634E-7 C. Participants note that the conducting shell affects the potential calculations at its inner surface and the outer surface of the insulating sphere, leading to confusion in the initial attempts. It is emphasized that the potential must account for the influence of the conducting shell, prompting users to revisit the definitions of electric potential. The conversation highlights the complexities involved in multi-layered electrostatic systems.
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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



|\vec{E}|=kQ_{enc.}/r^{2}
V=∫kQ_{enc.}/x^{2}dx

The Attempt at a Solution



1)E_{x}(P)= -33995.1 N/C
Q=\rhoV=-3.634E-7C
E=kQ/r^{2}

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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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).
 
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