Question

Two large conducting sheets are kept parallel to each other as shown in the figure. In equilibrium, the charge density on facing surfaces is ${\sigma }_1$ and ${\sigma }_2$. What is the net value of electric field at $\text{A}$ ?

• A. $\frac{{\sigma }_1}{{ϵ}_0}\hat{i}$
• B. $-\frac{{\sigma }_2}{{ϵ}_0}\hat{i}$
• C. $\frac{{\sigma }_1+{\sigma }_2}{2{ϵ}_0}\hat{i}$
• D. $\frac{{\sigma }_1-{\sigma }_2}{2{ϵ}_0}\hat{i}$

Answer 1

The correct option is D $\frac{{\sigma }_1-{\sigma }_2}{2{ϵ}_0}\hat{i}$


Electric field at point $\text{A}$ due to sheet $1$,

$\overrightarrow{E_1}=\frac{{\sigma }_1}{2{ϵ}_0}\hat{i}$

And due to sheet $2$,

$\overrightarrow{E_2}=\frac{{\sigma }_2}{2{ϵ}_0}\widehat{(-i)}=-\frac{{\sigma }_2}{2{ϵ}_0}\hat{i}$

So, net electric field at $\text{A},$

${\overrightarrow{E}}_{net}=\overrightarrow{E_1}+\overrightarrow{E_2}=\frac{{\sigma }_1-{\sigma }_2}{2{ϵ}_0}\hat{i}$

So, option $\left(d\right)$ is correct.