Question

Two charges of equal magnitudes kept at a distance r exert a force F on each other. If the charges

are halved and distance between them is doubled, then the new force acting on each charge is

(a) F/8

(b) F/4

(C) 4F

(d) F/16​

Answer 1

Explanation:

Original force,

F = k qq/2

New force,

F=k q/2 × q/2 ÷ (2r)^2

= 1/16 F

= F /16

Hence,

option (D) is correct answer.

Answer 2

Answer:

The new force acting on each charge is F/16 i.e.option(d).

Explanation:

From Columb's law, we have,

$F=\frac{kQ_1Q_2}{r^2}$        (1)

Where,

F=force on the charges

k=proportionality constant

Q₁, Q₂=charges

r=distance between the charges

From question,

∴Q₁=Q₂=Q                (charges of equal magnitudes)

The force acting is given as,

$F=\frac{k\timesQ\times Q}{r^2}=\frac{k\times Q^2}{r^2}$            (2)

Now if the charges are halved and the distance between them is doubled then,

$F'=\frac{k\times \frac{Q}{2}\times \frac{Q}{2}}{(2r)^2}=\frac{k\times Q^2}{16r^2}$      (3)

By comparing equations (2) and (3),

$F'=\frac{F}{16}$

Hence, the new force acting on each charge is F/16 i.e.option(d).