Question

Two spherical bodies, each of mass $50$ kg, are placed at a separation of $20$cm, Equal charges are placed on the bodies and it is found that the force of Coulomb repulsion equals the gravitational attraction in magnitude, The magnitude of the charge placed on either body is $x\times {10}^{-9}C$. Find the value of $x$.

Answer 1

mass $=50kg$
$r=20cm=0.2m$
$F_G=G\frac{m_1{m}_2}{r^2}=\frac{6.67\times {10}^{-11}\times 2500}{0.04}$
Coulomb's force $F_c=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r^2}=9\times {10}^9\frac{q^2}{0.04}$
Since, $F_G=F_C=\frac{6.7\times {10}^{-11}\times 2500}{0.04}=\frac{9\times {10}^9\times q^2}{0.04}$
$\Rightarrow q^2=\frac{6.7\times {10}^{-11}\times 2500}{0.04}=\frac{6.7\times {10}^{-9}}{9\times {10}^9}\times 25$
$=18.07\times {10}^{-18}$
$q=\sqrt{18.07\times {10}^{-18}}=4.3\times {10}^{-9}C$

So, the value of $x$ is $4.3$.