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. Find the magnitude of the charge placed on either body.

Answer 1

Given,

Mass = 50 kg,

r = 20 cm = 0.2 m

We know,

$F_G=G\frac{m_1{m}_2}{r^2}$

$=\frac{67\times {10}^{-11}\times 2500}{0.04}$

Coulomb's force,

$F_e=\frac{1}{4\pi {\in }_0}\frac{q_1{q}_2}{r^2}$

$=9\times {10}^9\frac{a^2}{0.04}$

Since $F_G=F_e$

$\Rightarrow \frac{6.7\times {10}^{-11}\times 2500}{0.04}=\frac{9\times {10}^9\times q^2}{0.04}$

$\Rightarrow q^2=\frac{6.7\times {10}^{-11}\times 2500}{9\times {10}^9}$

$=\frac{9\times {10}^9}{0.04}=1809\times {10}^{-18}$

$q=\sqrt{18,09\times {10}^{-18}}$

$=4.3\times {10}^{-9}C$