Question

Two pith balls, each of mass 1.8 g are suspended from the same point by silk threads each of length 20 cm. When equal charge $Q$ is given to both the balls, they separate until the two threads become perpendicular. Then the charge $Q$ on each pith ball is $\left[\frac{1}{4\pi \varepsilon_0 }=9 \times10^{9} N m^{2} C^{-2}\right]$

• A. $3 \times10^{-7} C$
• B. $4 \times10^{-7} C$
• C. $5 \times10^{-7} C$
• D. $2 \times10^{-7} C$

Answer 1

Answer: (B)

$4 \times10^{-7} C$

Answer 2

Given, mass of ball $m = {1.8 \,g = 1.8 \times 10^{-3} \, kg}$
Length of pendulum $l = {20 \, cm = 20 \times 10^{-2} \, m}$

From figure, at equilibrium
$\sum F_x = T \sin \theta - F_e =0$ ...(i)
$\sum F_y = T \cos \theta = mg = 0$ ....(ii)
From equation (ii),
$T = \frac{mg}{\cos \theta}$
Eliminating $T$ from equation (i) we get
$F_e = mg \, \tan \theta$
or $\frac{kq^2}{r^2} = mg \, \tan \theta ,$ where $k = \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9$
$\therefore \:\:\: \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{(2 l \, \sin \theta)^2} = mg \, \tan \theta$ $(\because \:\: r = 2l \sin \theta)$
or $q= \sqrt{16 \pi \varepsilon_0 l^2 mg \, \tan\, \theta \, \sin^2 \, \theta}$
Substituting the values of $l, \theta$ and $m$ we have
$q = \sqrt{4 \times4\pi \varepsilon_0 \, (20 \times 10^{-2} )^2 (1.8 \times 10^{-3})(10) \times \tan \, 45^\circ \, \sin^2 \, 45^\circ}$
$q = 4 \times 10^{-7} \, C$