Question

Two non-conducting infinite plane sheets having charges $Q$ and $2Q$ are placed parallel to each other as shown in figure. The charge distribution on four faces of two plates are also shown. The electric field intensities at three points $1,2$ and $3$ are ${\overrightarrow{E}}_1,{\overrightarrow{E}}_2$ and ${\overrightarrow{E}}_3$ repectively then the magnitudes of ${\overrightarrow{E}}_1,{\overrightarrow{E}}_2$ and ${\overrightarrow{E}}_3$ are respectively [Take $S$ is face area of plates]

• A. $\text{Zero};\frac{Q}{2{\epsilon }_{0}S};\text{zero}$
• B. $\frac{5Q}{6{\epsilon }_{0}S};\frac{Q}{{\epsilon }_{0}S};\frac{Q}{3{\epsilon }_{0}S}$
• C. $\frac{5Q}{6{\epsilon }_{0}S};\frac{Q}{2{\epsilon }_{0}S};\text{Zero}$
• D. $\text{Zero};\frac{Q}{{\epsilon }_{0}S};\text{Zero}$

Answer 1

The correct option is C $\frac{5Q}{6{\epsilon }_{0}S};\frac{Q}{2{\epsilon }_{0}S};\text{Zero}$

Let us assume, Rightward direction $+ve$ and Leftward direction $-ve$

Electric field due to charges on faces of non conducting plane sheets at point $\left(1\right)$ is as shown in the figure below.


For point $\left(1\right):$

${\overrightarrow{E}}_1=E_A-(E_B+E_C+E_D)$

$=\frac{1}{2{\epsilon }_{0}S}[\frac{2Q}{3}-\frac{Q}{3}-\frac{Q}{2}-\frac{3Q}{2}]$
$=\frac{1}{2{\epsilon }_{0}S}[\frac{2Q}{3}-\left(\frac{2+3+9}{6}\right)Q]=\frac{1}{2{\epsilon }_{0}S}[\frac{2Q}{3}-\frac{14}{6}Q]$

$\Rightarrow |\overrightarrow{E_1}|=\frac{5Q}{2{\epsilon }_{0}S}$

For point $\left(2\right):$

${\overrightarrow{E}}_2=(E_A+E_B)-(E_C+E_D)$

$\Rightarrow \overrightarrow{E_2}=\frac{1}{2{\epsilon }_{0}S}[\frac{2Q}{3}+\frac{Q}{3}-\frac{Q}{2}-\frac{3Q}{2}]=\frac{-Q}{2{\epsilon }_{0}S}$

$|{\overrightarrow{E}}_2|=\frac{Q}{2{\epsilon }_{0}S}$

For point $\left(3\right):$

${\overrightarrow{E}}_3=(E_A+E_B+E_C)-E_D$

$=\frac{1}{2{\epsilon }_{0}S}[\frac{2Q}{3}+\frac{Q}{3}+\frac{Q}{2}-\frac{3Q}{2}]$

$=\frac{1}{2{\epsilon }_{0}S}\left[\frac{4Q+2Q+3Q-9Q}{6}\right]$

$=0$

Hence, option (c) is the correct answer.