Question

Two equal charges are placed at a separation of $1\text{m}$. What should be the magnitude of the charges so that the force between them equals the weight of a $50\text{kg}$ person ?
(Take $g=10{\text{m/s}}^2$)

• A. $3\times {10}^{-4}C$
• B. $2.3\times {10}^{-4}C$
• C. $3.2\times {10}^{-4}C$
• D. $5\times {10}^{-5}C$

Answer 1

The correct option is B $2.3\times {10}^{-4}C$

Given that,
Distance between charges, $r=1\text{m}$
Weight of man, $W=mg=500\text{N}$

The Coulombic force is given by
$F_e=k\frac{q_1{q}_2}{r^2}$
where $q_1$ and $q_2$ are charge and $k$ is constant.
$\Rightarrow F_e=\frac{9\times {10}^9\times q^2}{(1{)}^2}(\because q_1=q_2=q)$
According to question,
$F_e=W$
$\Rightarrow 9\times {10}^9\times q^2=500$
$q^2=\frac{500}{9\times {10}^9}$
or, $q=\sqrt{\frac{50}{9}\times {10}^{-8}}$
$\Rightarrow q=\sqrt{\frac{50}{9}}\times {10}^{-4}$
or $q=\frac{5\sqrt{2}}{3}\times {10}^{-4}=1.66\times 1.414\times {10}^{-4}$
$\therefore q=2.34\times {10}^{-4}\text{C}$
Hence, option (b) is correct.