Question

Two identical copper spheres carrying charges +Q and -9Q separated by a certain distance has attractive force F. If the spheres are allowed to rouch each other and moved to distance of separation 'x'. So that the force between them becomes $\frac{4F}{9}$. Then X is equal to

• A. $d$
• B. $2d$
• C. $d/2$
• D. $4d$

Answer 1

Answer: (C)

$d/2$

Two identical copper sphere carrying charge $Q$ and $-9Q$ force between them $=\frac{4F}{q}$

$x=$ distance of separation

$\frac{4F}{q}=\frac{1}{4\pi {ϵ}_0}\frac{Q\times 9Q}{x^2}$

$x^2=\frac{9\times {10}^9\times {9Q}^2\times 9}{4F}$

or, $F$ being a attractive force

$x=d/2$ equating this the solution will be held.