Question

Two small identical balls P and Q, each of mass $\frac{\sqrt{3}}{10}$ gram, carry identical charges and are suspended by threads of equal lengths. At equilibrium, they position themselves as shown in Fig. 20.5 what is the charge on each ball. Given $\frac{1}{4\pi {\epsilon }_0}=9\times {10}^{9}N{m}^2{C}^{-2}$ and take $g=10N/Kg.$

• A. ${10}^{-3}C$
• B. ${10}^{-5}C$
• C. ${10}^{-7}C$
• D. ${10}^{-9}C$

Answer 1

The correct option is C

${10}^{-7}C$

Refer to Fig. 20.25. Let us consider forces on a ball, say, Q. Three forces act on it: (i) tension T in the thread, (ii) force mg due to gravity and (iii) force F due to Coluomb repulsion along +x-direction. For equilibrium, the sum of the x and y components of these forces must be zero, i.e,
$Tcos{60}^{\circ }-F=0$
and $Tsin{60}^{\circ }-mg=0$

These equations give F=mg cot ${60}^{\circ }=\frac{\sqrt{3}}{10}\times {10}^{-3}\times 10\times \frac{1}{\sqrt{3}}N$. Now
$F=\frac{1}{4\pi {ϵ}_0}.\frac{q^2}{r^2}$
Putting $F={10}^{-3}N,r=0.3m$
and $\frac{1}{4\pi {ϵ}_0}=9\times {10}^9$, we get $q={10}^{-7}$ coulomb.