Question

Two sphere of electric charges $+2$ nC and $-8$ nC are placed at a distance '$d$' apart. If they are allowed to touch each other, what is the new distance between them to get a repulsive force of same magnitude as before?

• A. $\frac{d}{2}$
• B. $d$
• C. $\frac{3d}{4}$
• D. $\frac{4d}{3}$

Answer 1

Answer: (C)

$\frac{3d}{4}$

Initially $2nC$ and $-8nC$ are separated by a distance $d$.

Electrostatic force acting between them initially $F=\frac{k{q}_1{q}_2}{d^2}$

$\therefore$ $F=\frac{\left(2\right)\left(8\right)k}{d^2}=\frac{16k}{d^2}$

Now the two charges are allowed to touch each other, they share equal charges among themselves.

Total charge of the system $Q=2-8=-6nC$

Thus new charge on each sphere $q_1^{\prime }=q_2^{\prime }=-3nC$

Let the new distance between them be $d^{\prime }$.

Thus force acting between them $F^{\prime }=\frac{k{q}_1^{\prime }{q}_2^{\prime }}{(d^{\prime }{)}^2}$

$\therefore$ $F^{\prime }=\frac{k(-3)(-3)}{(d^{\prime }{)}^2}=\frac{9k}{(d^{\prime }{)}^2}$

But $F^{\prime }=F$

$\therefore$ $\frac{9k}{(d^{\prime }{)}^2}=$ $\frac{16k}{d^2}$

Or $(d^{\prime }{)}^2=\frac{9}{16}{d}^2$

$⟹$ $d^{\prime }=\frac{3d}{4}$