Question

the electrostatic force between charges of 200 microcoulomb and 500 microcoulomb placed in free space is 5 gf find the distance between the two charges take g is equal to 10 metre per second square​

Answer 1

Answer:

=> The electrostatic force between charges of 200 microcoulomb and 500 microcoulomb placed in free space is 5 gf.

thus, charge $q_1$ = 200μC = 200 * 10^-6 C = 2 * 10^-4 C,

charge  $q_2$ = 500 μC =  500 * 10^-6 C = 5 * 10^-4 C,

=> The electrostatic force between these two charges:

F = 5 gf = 5 * 10^-3 kgf = 5 * 10^-2 N.

=>  The distance (r) between the two charges:

F = 1 / 4πε₀ * q1 q2 / r^2

5 * 10^-2 = 9 * 10^9 * 2 * 10^-4 * 5 * 10^-4 / r^2

r^2 = 9 * 10^9 * 2 * 10^-4 * 5 * 10^-4  / 5 * 10^-2

r^2 = 90,000/ 5

r^2 = 18000

r = 134.1 ≈ 1.34 * 10^2 m

Therefore, the distance between the two charges is 1.34 * 10^2 m.

Answer 2

Explanation:

the answer of this question is 1.34×10^-2