Question

Learning Goal: To understand the behavior ofthe electric field at the surface of a conductor, and itsrelationship to surface charge on the conductor.
A conductor is placed in an external electrostatic field. Theexternal field is uniform before the conductor is placed within it.The conductor is completely isolated from any source of current orcharge.

PART A)
Which of the following describes the electricfield inside this conductor?

It is in thesame direction as the original external field.
It is in theopposite direction from that of the original externalfield.
It has adirection determined entirely by the charge on itssurface.
It is alwayszero.
PART B)
The charge density inside theconductor is:

0
non-zero;but uniform
non-zero;non-uniform
infinite
PART C)
Assume that at some point just outside thesurface of the conductor, the electric field has magnitudeE and is directed toward thesurface of the conductor. What is the charge density eta on the surface of the conductor at thatpoint?
Express your answer in terms ofE and epsilon_0.
eta =

Answer 1

Complete Question

The complete question is shown on the first uploaded image  

Answer:

a   it is always zero

b  0

c  $\eta = -\epsilon _o E$

Explanation:ss

Here the  net charge is  on the outer surface of the conductor thus this means that the net charge inside the conductor is zero

Generally the charge density of a conductor is dependent on the charge per unit area  which implies that the charge density is dependent on the net charge  so this  means that the charge density inside the conductor is zero

 

Generally the direction of electric field this from the  positive charge to the negative charge  so from the question we can deduce  that the negative charge is located on the surface of the conductor

    So We can mathematically define the charge density on the surface of the electric field as

             ∮$E \cdot dA = \frac{-Q}{\epsilon _o}$

Where E is the electric field

          $dA$ change in unit area

           $-Q$ is the negative charge

          $\epsilon _o$  is the permittivity of free space

So

          $EA = \frac{-Q}{\epsilon _o }$

           $\frac{Q}{A} = -\epsilon _o E$

          $\eta = -\epsilon _o E$

Where $\eta$ is the charge density