Two spherical bodies, each of mass $50 \mathrm{~kg}$, are placed at a separation of $20 \mathrm{~cm}$. 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.
$F_{1}=\frac{G m_{1} m_{\Omega}}{r^{2}}$
$F_{1}=\frac{6.67 \times 10^{-11} \times 2500}{(0.2)^{2}}$
Coulomb's Force $F_{2}=\frac{k q_{1} q_{2}}{r^{2}}$
$F_{2}=\frac{9 \times 10^{9} \times q^{2}}{(0.2)^{2}}$
Now,
$F_{1}=F_{2}$
$\frac{6.67 \times 10^{-11} \times 2500}{0.04}=\frac{9 \times 10^{9} \times q^{2}}{0.04}$
$q^{2}=18.07 \times 10^{-18}$
$q=4.3 \times 10^{-9} \mathrm{C}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.