Question

Two point charges separated by a distance $d$ repel each with a force of $9N.$ If the separation between them becomes $3d,$ the force repulsion will be

• A. 1 N
• B. 3 N
• C. 6 N
• D. 27 N

Answer 1

The correct option is A 1 N

Let the two charges be $Q_1$ & $Q_2$ and according to coulom's force,

$\overrightarrow{F}=\frac{K{Q}_1{Q}_2}{R^2}$

where, ATQ, $\overrightarrow{F}=9N$

$R=d$

$\Rightarrow 9=\frac{K{Q}_1{Q}_2}{d^2}$

${9d}^2=K{Q}_1{Q}_2⟶\left(1\right)$

Now, in $2^{nd}$ case when $d\to 3d$, force is given by,

${\overrightarrow{F}}^1=\frac{K{Q}_1{Q}_2}{{\left(3d\right)}^2}=\frac{K{Q}_1{Q}_2}{9d^2}⟶\left(2\right)$

Substituting (2) into (1) will result to

${\overrightarrow{F}}^1=\frac{K{Q}_1{Q}_2}{K{Q}_1{Q}_2}=1N$

$\therefore$ Option (A) is the correct option.