Question

Two infinite sheets having charge densities ${\sigma }_1$ and ${\sigma }_2$ are placed in two perpendicular planes whose two-dimensional view is shown in the figure. The charges are distributed uniformly on the sheets in electrostatic equilibrium condition. Four points are marked $\text{I, II, III}$ and $\text{IV}$. The electric field intensities at these points are ${\overrightarrow{E}}_1,{\overrightarrow{E}}_2,{\overrightarrow{E}}_3$ and ${\overrightarrow{E}}_4$ respectively. The correct expression for the electric field intensities is

• A. $\left|{\overrightarrow{E}}_1\right|=\left|{\overrightarrow{E}}_2\right|=\left|{\overrightarrow{E}}_3\right|=\frac{\sqrt{{\sigma }_1^2+{\sigma }_2^2}}{2{\epsilon }_0}\ne \left|{\overrightarrow{E}}_4\right|$
  
• B. $\left|{\overrightarrow{E}}_2\right|=\left|{\overrightarrow{E}}_4\right|=\frac{\sqrt{{\sigma }_1^2+{\sigma }_2^2}}{2{\epsilon }_0}$
  
• C. $\left|{\overrightarrow{E}}_1\right|=\left|{\overrightarrow{E}}_2\right|=\left|{\overrightarrow{E}}_3\right|=\left|{\overrightarrow{E}}_4\right|=\frac{\sqrt{{\sigma }_1^2+{\sigma }_2^2}}{2{\epsilon }_0}$
  
• D. None of these

Answer 1

The correct option is C $\left|{\overrightarrow{E}}_1\right|=\left|{\overrightarrow{E}}_2\right|=\left|{\overrightarrow{E}}_3\right|=\left|{\overrightarrow{E}}_4\right|=\frac{\sqrt{{\sigma }_1^2+{\sigma }_2^2}}{2{\epsilon }_0}$


Electric field due $\overrightarrow{E}$ due to infinite sheet of charge density $\sigma$ is,$$E=\frac{\sigma }{2{\epsilon }_0}$$The directions of electric field at all the poins due the perpendicular sheet are shown below.


Net electric field at point $\text{I}$:

${\overrightarrow{E}}_{\text{I}}=\frac{{\sigma }_1}{2{\epsilon }_0}\hat{i}+\frac{{\sigma }_2}{2{\epsilon }_0}\hat{j}$

Net electric field at point $\text{II}$:

${\overrightarrow{E}}_{\text{II}}=\frac{{\sigma }_2}{2{\epsilon }_0}\hat{j}-\frac{{\sigma }_1}{2{\epsilon }_0}\hat{i}$

Net electric field at point $\text{III}$:

${\overrightarrow{E}}_{\text{III}}=-\frac{{\sigma }_1}{2{\epsilon }_0}\hat{i}-\frac{{\sigma }_2}{2{\epsilon }_0}\hat{j}$

Net electric field at point $\text{IV}$:

${\overrightarrow{E}}_{\text{IV}}=\frac{{\sigma }_1}{2{\epsilon }_0}\hat{i}-\frac{{\sigma }_2}{2{\epsilon }_0}\hat{j}$

So,
$\left|{\overrightarrow{E}}_{\text{I}}\right|=\left|{\overrightarrow{E}}_{\text{II}}\right|=\left|{\overrightarrow{E}}_{\text{III}}\right|=\left|{\overrightarrow{E}}_{\text{IV}}\right|=\sqrt{{\left(\frac{{\sigma }_1}{2{\epsilon }_0}\right)}^2+{\left(\frac{{\sigma }_2}{2{\epsilon }_0}\right)}^2}$

$\therefore \left|{\overrightarrow{E}}_{\text{I}}\right|=\left|{\overrightarrow{E}}_{\text{II}}\right|=\left|{\overrightarrow{E}}_{\text{III}}\right|=\left|{\overrightarrow{E}}_{\text{IV}}\right|=\frac{\sqrt{{\sigma }_1^2+{\sigma }_2^2}}{2{\epsilon }_0}$

Hence, option (c) is the correct answer.