Question

Two small identical balls $\text{P}$ and $\text{Q}$, each of mass $\frac{\sqrt{3}}{10}\text{gram}$, carrying identical charges are suspended by threads of equal lengths. At equilibrium, they position themselves as shown in the figure, what is the charge on each ball ?.
[ Given $\frac{1}{4\pi {ϵ}_0}=9\times {10}^9{\text{Nm}}^2{\text{C}}^{-2}$ and take $g=10{\text{ms}}^{-2}$.]

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

Answer 1

The correct option is C ${10}^{-7}\text{C}$

Given, Mass of ball $m=\frac{\sqrt{3}}{10}\text{grams}=\sqrt{3}\times {10}^{-4}\text{kg}$

$g=10{\text{ms}}^{-2}$


Refer to figure, let us consider forces on a ball, at point $\text{Q}$.

Three forces act on it:
$\text{1}.$ Tension $T$ in the thread,
$\text{2}.$ Force $mg$ due to gravity and
$\text{3}.$ Force $F$ due to Coulomb repulsion lets say along $+x$ direction.

For equilibrium, the sum of the $x$ and $y$ components of these forces must be zero, i.e.

$T\cos {60}^{\circ }-F=0.........\left(1\right)$

And

$T\sin {60}^{\circ }-mg=0........\left(2\right)$

On solving equation $\left(1\right)$ and $\left(2\right)$ we get,

$F=mg\cot {60}^{\circ }$

Substituting the given values in this we get,

$F=\sqrt{3}\times {10}^{-4}\times 10\times \frac{1}{\sqrt{3}}$

$\therefore F={10}^{-3}\text{N}$

Now from coulomb's law of force,

$F=\frac{1}{4\pi {ϵ}_0}\frac{q^2}{r^2}$

Putting, $F={10}^{-3}\text{N},r=0.3\text{m}$

And $\frac{1}{4\pi {ϵ}_0}=9\times {10}^9$, we get

${10}^{-3}=9\times {10}^9\times \frac{q^2}{{0.3}^2}$

$\therefore q={10}^{-7}\text{C}$.

Hence, option (c) is correct option.