Question

Two identical copper spheres carrying chatges $+Q$ and $-9Q$ separated by a certain distance has attractive force F. If the spheres are allowed to touch eacxh 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

The correct option is C d/2

Two identical copper sphere carrying charges $+Q$ and $-9Q$

Force between them $=\frac{4F}{q}$

$x=$ distance of separation $x$.

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

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

$\because$ $F$ being attractive force

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