Question

Two spherical conductors A and B having equal radii and carrying equal charges in them repel each other with a force F when kept apart at some distance. A third spherical conductor C having same radius as that of A but uncharged is bought in contact with B and then with A. then new force of repulsion between B and A will be :

• A. $F$
• B. $\frac{3F}{4}$
• C. $\frac{F}{8}$
• D. $\frac{3F}{8}$

Answer 1

The correct option is D $\frac{3F}{8}$

As we know that $F\propto \frac{q^2}{d^2}$
When B is in contact with C; charge on B and C
$\frac{q+0}{2}=\frac{q}{2}$
When A is in contact with C; charge on A and C
$\frac{q+\frac{q}{2}}{2}=\frac{3q}{4}$
$\therefore$
Charge on $B=\frac{q}{2}$Charge on $A=\frac{3q}{4}$
Force between B and A
$F_1\propto \frac{\frac{q}{2}\cdot \frac{3q}{2}}{d^2}$ ....(ii)
From Eqs. (i) and (ii),
$\frac{F_1}{F}=\frac{3q^2/8d^2}{q^2/d^2}=\frac{3}{8}$
$F_1=\frac{3}{8}F$