Question

Two identical small balls of equal charge q & equal mass m = 0.9 kg are placed at a separation of r= 1 min
air on a rough surface. if coefficient of static friction is u.= 0.9. Find the magnitude of maximum charge q such
that both balls remain at rest.
(1) + 30 uC
(2) 90 uc
(3) + 60 uc
(4) +20 uc​

Answer 1

Answer:

Option (1) 30 μC

Explanation:

By Coulomb's law the force on the two charges q₁ and q₂, separated by a distance r is given by

$\boxed{F=k\frac{q_1q_2}{r^2}}$

Where k is Coulom's law constant, k = 9 × 10⁹ for air

Here the charges on the ball are same = q

The distance between the balls = r

Thus, the force acting between them

$F=k\frac{q^2}{r^2}$

The Frictional force acting between the charges = μmg

                                                                                = 0.9 × 0.9 × 10

                                                                                = 8.1 N

Both the balls will remain at rest till

$k\frac{q^2}{r^2}=8.1$

$\implies 9\times 10^9\times \frac{q^2}{1^2}=8.1$

$\implies  q^2=\frac{8.1}{9\times 10^9}$

$\implies  q^2=9\times 10^{-10}$

$\implies  q=3\times 10^{-5}=30\times 10^{-6} \text{ C}=30\mu\text{C}$