Question

Two equal and like charges when placed 5 cm apart experience a repulsive force of 0.144 newton. the magnitude of the charge in micro-coulomb will be.

Answer 1

Hello Dear.

Here is the answer---

Given ⇒

Distance between the the charges = 5 cm.
 = 0.05 m.
Repulsive Force between two charges = 0.144 N.

Let the two charges be a.

Now, According to the coulomb law, 

Repulsive Force(F) = (K × q₁ × q₂) ÷ d²

We know, k = 8.99 × 10⁹ Nm²/C²                   
   = 9 × 10⁹ Nm²/C²   [Approx]

Thus, Putting the above values in the Question,

 0.144 = (9 × 10⁹ × a × a) ÷ (0.05)²
0.00036 = 9 × 10⁹ × a²
  ⇒ a = 2 μC.


Thus, the magnitude of the charge will be 2 μC.

Hope it helps.