Question

28. Electric field due to an infinite non-conducting sheet of surface charge density Sigma, at a distance r from it is
(1)Sigma/2E⁰
(2)Sigma/E⁰
(3) Sigma²/2E⁰
(4)Sigma²/E⁰​

Answer 1

Question:

Electric field due to an infinite non-conducting sheet of surface charge density $\sigma$ , at a distance r from it is

• $\frac{\sigma}{2 \epsilon_0}$
• $\frac{\sigma}{\epsilon_0}$
• $\frac{\sigma^2}{2 \epsilon_0}$
• $\frac{\sigma^2}{ \epsilon_0}$

Solution:

(See the attachment for reference)

Here we have:

• $\sigma = charge ~density$
• $dE= Electric~ field~at~P$

Breaking dE in components we have $dEsin \theta$ and $dE cos \theta$

We have

$\sf \sum q = \sigma S \\\\ \sf \sf By~flux,~ we ~have ~, \\\\ \phi E = \oint_S \vec{E} \vec{d}s \\\\ \phi E = \oint_{S_1}\vec{E} \vec{d}s + \oint_{S_2}\vec{E} \vec{d}s+ \oint_{S_3} \vec{E} \vec{d}s \\\\ \oint_{S_1} Eds~cos 0^{\circ}+\oint_{S_2} Eds~cos0^{\circ} +\oint_{S_3} Eds~cos90^{\circ} \\\\ E \oint_{S1} ~dS + E \oint_{S_2} ~dS \\\\ ES+ES = 2ES - - - - [i]$

By, Gauss's law, we have:

$\phi E = \frac{ \sum q} {\epsilon_0} \implies \frac{ \sigma S} {\epsilon_0} - - - - [ii]$

From [i] and [ii], we have,

$\phi E= \phi E \\\\ 2ES= \frac{ \sigma S} {\epsilon_0} \\\\ 2E \cancel{S}= \frac{ \sigma \cancel{ S} } {\epsilon_0} \\\\ \boxed{E= \frac{\sigma}{2 \epsilon_0}}$

Hence, the correct answer is A.