Question

Two identical metal spheres with charges $2Q$ and $-Q$ are separated by a distance. Force between them is F. They are connected by a conducting wire and then removed. Force between them now is

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

Answer 1

Answer: (B)

$\frac{F}{8}$

Answer 2

Force between metal spheres, initially
\hspace30mm F = K \frac{(2Q)(-Q)}{r^2}
When the metal spheres are connected by a conducting wire, charge flows from one to other, until both are at common potential. As the spheres are identical, therefore, charge on them becomes, $q_1 = q_2 = \frac{2Q + (-Q)}{2} = \frac{Q}{2}$
$\therefore \, \, F' = \frac{K\left(\frac{Q}{2}\right)\left(\frac{Q}{2}\right)}{r^{2}} = \frac{-F}{8}$
(Negative sign shows that the direction of force has changed).