Question

Electric field intensity at a point outside uniformly charged thin infinite plane sheet is \(E_1\). The electric field intensity at a point near and outside the surface of a positively charged conductor of any shape is \(E_2\). The relation between magnitude of \(E_1\) and \(E_2\) is (assume air as the medium)

• A. \(E_1 = E_2\)
• B. \(2E_1 = E_2\)
• C. \(E_1 = 2E_2\)
• D. \(E_1 = 4E_2\)

Hint:

Electric field near a conductor is always twice the field due to a non-conducting infinite plane sheet having the same surface charge density.

Answer 1

The Correct Option is B

Step 1: Electric field due to an infinite plane sheet.
For a uniformly charged infinite plane sheet, the electric field outside the sheet is given by:
\[ E_1 = \frac{\sigma}{2\varepsilon_0} \] where \(\sigma\) is surface charge density.

Step 2: Electric field near a charged conductor.
For a charged conductor, the electric field just outside the surface is:
\[ E_2 = \frac{\sigma}{\varepsilon_0} \]
Step 3: Compare the two expressions.
\[ E_2 = 2E_1 \]
Step 4: Conclusion.
Hence, the correct relation is \(2E_1 = E_2\).