Question

Two spheres of electric charges $+2nC$ and $-8nC$ are placed at a distance ${}^{\prime }{d}^{\prime }$-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. $\frac{4d}{3}$
• B. $\frac{3d}{4}$
• C. $d$
• D. $\frac{d}{2}$

Answer 1

Answer: (B)

In the first condition,
Given, $q_1=+2nC=2\times 10^{-9}C$
$q_2=-8nC$
$=-8\times 10^{-9}C$
$F=\frac{k{q}_1{q}_2}{r^2}$
$F=\frac{k\left(2\times 10^{-9}\right)\times \left(8\times 10^{-9}\right)}{d^2}$
$F=\frac{k\left(16\times 10^{-18}\right)}{d^2}\dots (i)$
In the second condition,
$F_1=\frac{k{q}_1{q}_2}{d^{\prime }2}$
After touching of sphere to each other the total charge $=2-8=-6nC$
Charge on each sphere $=3nC$
$=3\times 10^{-9}C$
$F_1=\frac{k\left(3\times 10^{-9}\right)\left(3\times 10^{-9}\right)}{{\left(d^{\prime }\right)}^2}$
$F_1=\frac{k\left(9\times 10^{-18}\right)}{{\left(d^{\prime }\right)}^2}$
From Eqs. (i) and (ii), we get
$\frac{k\left(16\times 10^{-18}\right)}{d^2}=\frac{k\left(9\times 10^{-18}\right)}{{\left(d^{\prime }\right)}^2}$
$\frac{d^2}{d^2}=\frac{16\times 10^{-18}}{9\times 10^{-18}}$
$\frac{d^2}{d^{\prime }2}=\frac{16}{9}$
$\frac{d}{d^{\prime }}=\sqrt{\frac{16}{9}}$
$\frac{d}{d^{\prime }}=\frac{4}{3}{d}^{\prime }$
$=\frac{4}{3}{d}^{\prime }$
$d^{\prime }=\frac{3}{4}d$