Question

Two-point charges A and B, having charges +Q and -Q respectively, are placed at a certain distance apart and the force acting between them is F. If 25% charge of A is transferred to B, then the force between the charges becomes

• A. $\frac{16F}{9}$
• B. $\frac{9F}{16}$
• C. $\frac{4F}{3}$
• D. F

Answer 1

The correct option is B

$\frac{9F}{16}$

Force:

A force is an influence that has the power to alter an object's motion. An object with mass can change its velocity, or accelerate, as a result of a force.

Given Data

Two-point charges A and B, having charges +Q and -Q respectively, are placed at a certain distance apart, and the force acting between them is F.

Formula used

$F=\frac{K{Q}^2}{r^2}$

The correct option (B)

If 25% of the charge of A transferred to B then

$q_A=Q-\frac{Q}{4}=\frac{3Q}{4}$and $q_B=-Q+\frac{Q}{4}=\frac{-3Q}{4}$

Force can be given according to the requirement $F_1=\frac{k{q}_A{q}_B}{r^2}$

By putting the value in this:

$F_1=\frac{k{\left({\displaystyle \frac{3Q}{4}}\right)}^2}{r^2}$

$F_1=\frac{9}{16}\left(\frac{KQ}{r^2}\right)$

$F_1=\frac{9F}{16}$

Thus the correct option is B.

The incorrect options (A), (C), (D)

From the analysis, it is clear all three options are incorrect.

Therefore the correct option is B.