Question

$Q_1,Q_2$ charges are separated by distance $‘d^{\prime }$. $F$ is the force between them. Now the value of $Q_2$ is halved. To have the same force between the charges, the distance of separation should be :

• A. $d$
• B. $\frac{d}{2}$
• C. $\frac{d}{\sqrt{2}}$
• D. $2d$

Answer 1

Answer: (C)

$\frac{d}{\sqrt{2}}$

Initially force between them $F=\frac{K{Q}_1{Q}_2}{d^2}$
Now $Q_2$ is halved so to have the same force let the distance be X

$\Rightarrow \frac{K\left(\frac{Q_2}{2}\right)Q_1}{X^2}=\frac{K{Q}_1{Q}_2}{d^2}$

$\Rightarrow \frac{1}{2X^2}=\frac{1}{d^2}$

$\Rightarrow \frac{1}{\sqrt{2}X}=\frac{1}{d}$

$\Rightarrow X=\frac{d}{\sqrt{2}}$