Two point charges Q1 and Q2 exert a force F on each other when kept certain distance apart. If the charge on each particle is halved and the distance between the two particles is doubled, then the new force between the two particles would be

Question

Two point charges Q1 and Q2 exert a force F on each other when kept certain distance apart. If the charge on each particle is halved and the distance between the two particles is doubled, then the new force between the two particles would be (A) (F/2) (B) (F/4) (C) (F/8) (D) (F/16)

Tardigrade Admin · Accepted Answer

Given F=(K Q1 Q2/r2) ...(i) if, Q1'=(Q1/2) Q2'=(Q2/2) r=2 r Then =F1'=(F/16)

Q. Two point charges $Q_{1}$ and $Q_{2}$ exert a force $F$ on each other when kept certain distance apart. If the charge on each particle is halved and the distance between the two particles is doubled, then the new force between the two particles would be

$\frac{F}{2}$

$\frac{F}{4}$

$\frac{F}{8}$

$\frac{F}{16}$

Given
$F = \frac{K Q _{1} Q _{2}}{r ^{2}} ... (i)$
if, $Q_{1}^{'} = \frac{Q _{1}}{2}$
$Q_{2}^{'} = \frac{Q _{2}}{2}$
$& r = 2 r$
Then $= F_{1}^{'} = \frac{F}{16}$

Q. Two point charges Q1​ and Q2​ exert a force F on each other when kept certain distance apart. If the charge on each particle is halved and the distance between the two particles is doubled, then the new force between the two particles would be

2F​

4F​

8F​

16F​