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
$\left[\frac{1}{4\pi {\epsilon }_0}=9\times 10^{9}N{m}^2{C}^{-2}\right]$

• A. $3\times 10^{-7}C$
• B. $4\times 10^{-7}C$
• C. $5\times 10^{-7}C$
• D. $2\times 10^{-7}C$

Answer 1

Answer: (B)

Given, mass of ball $m=1\cdot 8g=1\cdot 8\times 10^{-3}kg$
Length of pendulum $l=20cm=20\times 10^{-2}m$
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 $\frac{k{q}^2}{r^2}=mg\tan \theta ,$ where $k=\frac{1}{4\pi {\epsilon }_0}=9\times 10^9$
$\therefore \frac{1}{4\pi {\epsilon }_0}\frac{q^2}{(2l\sin \theta {)}^2}=mg\tan \theta$ $(\because r=2l\sin \theta )$
or $q=\sqrt{16\pi {\epsilon }_0{l}^{2}mg\tan \theta {\sin }^2\theta }$
Substituting the values of $l,\theta$ and $m$ we have
$q=\sqrt{4\times 4\pi {\epsilon }_0(20\times 10^{-2}{)}^2(1.8\times 10^{-3})(10)\times \tan 45^{\circ }{\sin }^{2}45^{\circ }}$
$q=4\times 10^{-7}C$