Question

Two charges $Q$ & $-Q$ are kept at some distance, so that electric field at the mid point is $\overrightarrow{E}$. Find the force on $Q$ due to the field of $-Q$. (charges are isolated)

• A. $\frac{Q\overrightarrow{E}}{2}$
• B. $\frac{Q\overrightarrow{E}}{4}$
• C. $4Q\overrightarrow{E}$
• D. $\frac{Q\overrightarrow{E}}{8}$

Answer 1

The correct option is D $\frac{Q\overrightarrow{E}}{8}$


Let $E_{+}$ be electric field at mid point due to positive charge,
$E_{-}$ be electric field at mid point due to negative charge

Let distance between the charges be $r$.

The electric field at the midpoint,
$\overrightarrow{E}={\overrightarrow{E}}_{+}+{\overrightarrow{E}}_{-}$
$=\frac{kQ}{(r/2{)}^2}\hat{i}+\frac{kQ}{(r/2{)}^2}\hat{i}=\frac{8kQ}{r^2}\hat{i}$
[taking positive x axis in rightward direction]

At position $Q$, the field due to $-Q$ is ${\overrightarrow{E}}_Q=\frac{kQ}{r^2}\hat{i}=\frac{\overrightarrow{E}}{8}$

Therefore, the force on charge $Q$ is $\overrightarrow{F}=\frac{Q\overrightarrow{E}}{8}$