Question

Two conducting plates $x$ and $y$ each of area $A$ are placed parallel to each other at a small separation $d.$ $X$ is given a charge $3q$ and $Y$ is given a charge $q.$ The potential difference between the plates is

• A. $\frac{qd}{2{\epsilon }_{0}A}$
• B. $\frac{qd}{{\epsilon }_{0}A}$
• C. $\frac{3qd}{2{\epsilon }_{0}A}$
• D. $\frac{2qd}{{\epsilon }_{0}A}$

Answer 1

Answer: (B)

${\sigma }_1=\frac{3q}{A},{\sigma }_2=\frac{q}{A}$

A : Area of plate

$\therefore$ Net electric field in between plates is given by,

$\overrightarrow{E}=\frac{{\sigma }_1}{2{\epsilon }_0}-\frac{{\sigma }_2}{2{\epsilon }_0}$

$=\frac{1}{2{\epsilon }_0}\left[\frac{3q-q}{A}\right]=\frac{2q}{2{\epsilon }_{0}A}=\frac{q}{{\epsilon }_{0}A}$

Now, $\overrightarrow{E}=\frac{U}{d}$

$\Rightarrow U=\overrightarrow{E}.d=\frac{qd}{{\epsilon }_{0}A}$ {d : distance between plate}

$\therefore$ Potential difference between two plates will be given by $=\frac{qd}{{\epsilon }_{0}A}$

$\Rightarrow$ Option (B) is the correct answer.