Question

Calculate the force of repulsion acting between two positive charges of magnitude 20 esu and 30 esu when they are kept at a distance of 2 centimetre in air.
Please solve...​

Answer 1

Answer:

Explanation:

we know that =

F = K $\frac{q_1q_2}{r^2}$

k = 9 ×$10^9$

$q_1 = 20 esu$ = 6.671281903964E-9 Coulombs

$q_2 = 30esu$ = 1.0006922855946E-8 Coulombs

r = 2cm = 0.02 m.

put the values..

F=  9 ×$10^9$ 6.671E-9×.000692E-8 / 0.02²

F=   9 ×$10^9$ × 1.154083e-16

F = 674.734306416 N

F = 674.7N