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damaskus [11]
4 years ago
5

To understand the behavior of the electric field at the surface of a conductor, and its relationship to surface charge on the co

nductor. A conductor is placed in an external electrostatic field. The external field is uniform before the conductor is placed within it. The conductor is completely isolated from any source of current or charge.
1) Which of the following describes the electricfield inside this conductor?
A) It is in the same direction as the original external field.
B) It is in the opposite direction from that of the original external field.
C) It has a direction determined entirely by the charge on its surface.
D) It is always zero.
2) The charge density inside the conductor is:______.
a) 0.
b) non-zero;but uniform.
c) non-zero;non-uniform.
d) infinite.
3) Assume that at some point just outside the surface of the conductor, the electric field has magnitudeE and is directed toward the surface of the conductor. What is the charge density eta on the surface of the conductor at that point?
Physics
1 answer:
ANEK [815]4 years ago
4 0

Answer:

1) Option D is correct.

The electric field inside a conductor is always zero.

2) Option A is correct.

The charge density inside the conductor is 0.

3) Charge density on the surface of the conductor at that point = η = -E ε₀

Explanation:

1) The electric field is zero inside a conductor. Any excess charge resides entirely on the surface or surfaces of a conductor.

Assuming the net electric field wasn't zero, current would flow inside the conductor and this would build up charges on the exterior of the conductor. These charges would oppose the field, ultimately (in a few nanoseconds for a metal) canceling the field to zero.

2) Since there are no charges inside a conductor (they all reside on the surface), it is logical that the charge density inside the conductor is also 0.

3) Surface Charge density = η = (q/A)

But electric field is given as

E = (-q/2πε₀r²)

q = -E (2πε₀r²)

η = (q/A) = -E (2πε₀r²)/A

For an elemental point on the surface,

A = 2πrl = 2πr²

So,

η = -E ε₀

Hope this Helps!!!

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