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 coulamb will be

Answer 1

Let magnitude of each charge is q

Given,
repulsive force , F = 0.144N
Separation between charges, r = 5cm = 0.05 m

∵ F = KQ₁Q₂/r² [ from Coulombs law ]
∴ 0.144 = 9 × 10⁹ × q × q/(0.05)²
⇒ 0.144 × (0.05)² = 9 × 10⁹ × q²
⇒ 0.144 × (0.05)²/9 × 10⁹ = q²
⇒144 × (0.05)² × 10⁻¹²/9 = q²
taking square root both sides,
⇒ 12 × 0.05 × 10⁻⁶/3 = q
⇒ 0.2 μC = q

Hence, magnitude of each charge = 0.2 μC

Answer 2

Hello Dear.

Here is your answer---

Given Conditions---

 Distance between two like charges = 5 cm.
                                                          = 0.05 m.

Repulsive force between the two charges = 0.144 Newton.

Let the two charges be a.

As per as Coulomb law, Using the Formula,

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.

Have a Marvelous Day.