Question

Two identical metallic spheres A and B carry charges $+Q$ and $-2Q$ respectively. The force between them is 'F' newton, when they are separated by a distance 'd' in air. The spheres are allowed to touch each other and are moved back to their initial position. The force between them now is:

• A. $2F$
• B. $\frac{F}{8}$
• C. $\frac{F}{4}$
• D. $8F$

Answer 1

The correct option is B $\frac{F}{8}$

intial force $F_i=\frac{k\left(Q\right)\left(\right(2Q)}{d^2}$
$F_i=\frac{2k{Q}^2}{d^2}$
Now when spheres are touched, charge distribution changes to make potential equal on both spheres
$\therefore Q_f=\frac{Q}{2}$ (as both the spheres are identical)

$F_f=\frac{k\left(\frac{Q}{2}\right)\left(\frac{Q}{2}\right)}{d^2}$

$F_f=\frac{k{Q}^2}{4d^2}$

$F_f=\frac{F}{8}$