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 uniformly charged infinite thin plane sheet 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. $E_1 = 4E_2$

Answer 1

Answer: (C)

Electric field intensity at a point near and outside the surface of any charged conductor is given by
$E_{1}=\frac{\sigma}{\varepsilon_{0}}\,\,\,\,\,..(i)$
where, $\sigma$ is charge density (charge per unit area) at a point.
Electric field intensity due to uniformly charged infinite thin plane sheet is given by
$E_{2}=\frac{\sigma}{2 \varepsilon_{0}}\,\,\,\,\,\,..(ii)$
From Eqs. (i) and (ii), we have
$E_{1}=\frac{2}{2} \times \frac{\sigma}{\varepsilon_{0}}=2 \times E_{2} \Rightarrow E_{1}=2 E_{2}$