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

Mass = 50 kg
Separation between the masses, r = 20 cm = 0.2 m
Let the change on each sphere be q.
Now, gravitational force, \[F_G = G\frac{\text{ m } _1 \text{ m } _2}{r^2}\] 

\[= \frac{6.7 \times {10}^{- 11} \times \left( 50 \right)^2}{\left( 0 . 2 \right)^2}\]

\[ = \frac{6.7 \times {10}^{- 11} \times 2500}{0 . 04}\]

\[\text {Coulomb force}, \text {F_c} = \frac{1}{4\pi \in_0}\frac{q_1 q_2}{r^2}\]

\[ = 9 \times {10}^9 \frac{q^2}{0 . 04}\]

Since F_G = F_c, we have :

\[\frac{6 . 7 \times {10}^{- 11} \times 2500}{0 . 04} = \frac{9 \times {10}^9 \times q^2}{0 . 04}\]

\[ \Rightarrow q^2 = \frac{6 . 7 \times {10}^{- 11} \times 2500}{9 \times {10}^9}\]

\[ = \frac{9 \times {10}^9}{0 . 04} = 1809 \times {10}^{- 18} \]

\[ \therefore q = \sqrt{18 . 09 \times {10}^{- 18}}\]

\[ = 4 . 3 \times {10}^{- 9} C\]

Thus, the charge of the spherical body is \[4 . 3 \times {10}^{- 9} C\]