Question

The force of repulsion between two identical positive charges is 0.800 N when the charges are 0.100 m apart. Find the value of each charge.

Answer 1

Answer:

Value of each charge is given as

$q = 9.43 \times 10^{-7} C$

Explanation:

As we know that electrostatic force between two charges is given as

$F = \frac{kq_1 q_2}{r^2}$

here we know that

$q_1 = q_2 = q$

$F = 0.800 N$

r = 0.100 m

now we have

$F = \frac{(9\times 10^9) q^2}{(0.100)^2}$

$0.800 = 9\times 10^{11} q^2$

$q = 9.43 \times 10^{-7} C$