Question

Two particles $\text{A}$ and $\text{B}$ $(\text{B}$ is to the right of $\text{A})$ having charges $8\times {10}^{-6}\text{C}$ and $-2\times {10}^{-6}\text{C}$, respectively, are held fixed with separation of $20\text{cm}$. Where should a third charged particle be placed so that it does not experience a net electric force.

• A. $5\text{cm}$ right of $\text{B}$
• B. $5\text{cm}$ left of $\text{A}$
• C. $20\text{cm}$ left of $\text{A}$
• D. $20\text{cm}$ right of $\text{B}$

Answer 1

The correct option is D $20\text{cm}$ right of $\text{B}$


Let the third charge $Q$ be placed at a distance $x$ to the right of $B$.

Then the net electric force acting on the charge $Q$,

${\overrightarrow{F}}_{net}={\overrightarrow{F}}_A+{\overrightarrow{F}}_B$

From the data given in the question,

$F_{net}=\frac{KQ\times 8\times {10}^{-6}}{(20+x{)}^2}-\frac{KQ(2\times {10}^{-6})}{x^2}$

Since the charge does not experience any net force, so

${\overrightarrow{F}}_{net}=0$

$\Rightarrow \frac{KQ\times 8\times {10}^{-6}}{(20+x{)}^2}-\frac{KQ(2\times {10}^{-6})}{x^2}=0$

$\Rightarrow 4x^2=(20+x{)}^2$

$\Rightarrow 20+x=\pm 2x$

On solving, we get

$x=20\text{cm}$

Hence, the charge $Q$ is placed at $20\text{cm}$ right of $\text{B}$.

Thus, option (d) is the correct answer.