Question

Two charges Q1, and Q2, are separated by a distance D. If both Q1, and D are doubled, the force is
(1) Increased by a factor of 2
(2) Decreased by a factor of 2
(3) Unchanged
(4) Increased by a factor of 4​

Answer 1

$\underline{ \underline{\huge{ \bf \pink{Given:-}}}}$

Charges = $Q_1\: and \:Q_2$

Distance = D

$\underline{ \underline{\huge{ \bf \green{To \:Find:-}}}}$

Change on force by factor

$\underline{ \underline{\huge{ \bf \orange{Solution:-}}}}$

$\boxed{ \sf{Force = \frac{k q_1.q_2}{r^2}}}$

$\therefore Original \:Force = \frac{kQ_1.Q_2}{D^2}$

$\large\dag$Since both Q1 and D are doubled,

$\large\bull$New Charge are $2Q_1 \:and \:Q_2$

$\large\bull$Also new Distance is 2D

New Force =$\frac{k2Q_1.Q_2}{(2D)^2}$

$\:\implies \frac{\cancel2kQ_1.Q_2}{\cancel4D^2}$

$\:\implies \frac{kQ_1.Q_2}{2D^2}$

$\:\implies\frac{1}{2}×(\frac{kQ_1.Q_2}{D^2})$

$\:\implies \frac{1}{2} × Original \:Force$

$\pink\therefore$Force is decreased by a factor of 2

$\large\green \maltese$ (2) is Correct.