Question

Two non conducting infinite plane sheets having charges Q and 2Q are placed parallel to each other as shown in Fig. The charge distribution on the four faces of the two plates is also shown. The electric field intensities at three points 1, 2, and 3 are ${\overrightarrow{E}}_1,{\overrightarrow{E}}_2,and{\overrightarrow{E}}_3$, respectively. Then the magnitudes of ${\overrightarrow{E}}_1,{\overrightarrow{E}}_3$ and ${\overrightarrow{E}}_2$ are respectively (surface area of plates):

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

Answer 1

The correct option is B $\frac{5Q}{6{\epsilon }_{0}S},\frac{Q}{2{\epsilon }_{0}S},0$

Let us take rightward as positive and leftward as negative.
These four faces can be considered four thin infinite plane sheets.
${\overrightarrow{E}}_1=\frac{1}{2{\epsilon }_{0}S}[\frac{2Q}{3}-\frac{Q}{3}-\frac{Q}{2}-\frac{3Q}{2}]=\frac{-5Q}{6{\epsilon }_{0}S}$
${\overrightarrow{E}}_3=\frac{1}{2{\epsilon }_{0}S}[\frac{2Q}{3}+\frac{Q}{3}-\frac{Q}{2}-\frac{3Q}{2}]=\frac{-Q}{2{\epsilon }_{0}S}$
${\overrightarrow{E}}_2=\frac{1}{2{\epsilon }_{0}S}[\frac{2Q}{3}+\frac{Q}{3}+\frac{Q}{2}-\frac{3Q}{2}]=0$