Question

Assuming that two 1 kg balls have equal magnitude charges, how much net charge would each need to have in order for the electric force between the balls to have the same magnitude as the gravitational force betwee n the balls? If you want the balls to remain fixed in place without moving, should they have the same sign of charge or opposite signs?

Answer 1

Answer:

$q=8.6 \times 10^{-11} C$

Explanation:

mass of each ball, m = 1 kg

Let the charge on each ball is q.

If the balls remains fix then the gravitational force is balanced by the electrostatic force between them and as the gravitational force is attractive in nature then the electrostatic force should be repulsive in nature.

The charges should be of same sign to get the electrostatic force is repulsive in nature.

Let the distance between the two balls is d.

The electrostatic force between them is given by

$F_{e}=\frac{kq^{2}}{d^{2}}$    ... (1)

The gravitational force between the two balls is given by

$F_{g}=\frac{Gm^{2}}{d^{2}}$    ... (2)

according to the question, gravitational force is equal to the electrostatic force, so by equation (1) and (2) ,we get

$\frac{kq^{2}}{d^{2}}=\frac{Gm^{2}}{d^{2}}$

$kq^{2}=Gm^{2}$

$9 \times 10^{9}q^{2}=6.67 \times 10^{-11}\times 1\times 1$

$q^{2}=7.41 \times 10^{-21}$

$q=8.6 \times 10^{-11} C$

Thus, the charge on each ball is $q=8.6 \times 10^{-11} C$