Question

The electric field intensity at a point near and outside the surface of a charged conductor of any shape is \(E_{1}\). The electric field intensity due to a uniformly charged infinite thin plane sheet having the same surface charge density is \(E_{2}\). The relation between \(E_{1}\) and \(E_{2}\) is

• A. \(2E_{1}=E_{2}\)
• B. \(E_{1}=E_{2}\)
• C. \(E_{1}=2E_{2}\)
• D. \(E_{1}=4E_{2}\)

Hint:

Just outside a conductor the field is sigma/epsilon0, while an infinite sheet gives sigma/2epsilon0.

Answer 1

The Correct Option is C

Step 1: Why the Conductor Differs
Inside a conductor the field is zero, so the flux passes out through one face of the Gaussian pillbox only: $E_{1}A=\dfrac{\sigma A}{\varepsilon_{0}}$.

Step 2: Why the Sheet Differs
A thin sheet has field on both sides, so flux passes through two faces: $2E_{2}A=\dfrac{\sigma A}{\varepsilon_{0}}$.

Step 3: Compare
Dividing gives $E_{1}=2E_{2}$.

Final Answer:
The conductor's surface field is twice the sheet's field. \[ \boxed{\text{(C) }E_{1}=2E_{2}} \]