Question

Two identical beads, each have a mass $m$ and charge $q$. When placed at a distance $R$ apart in a hemispherical bowl of radius $R$ with frictionless, non-conducting walls, the beads are at equilibrium as shown in figure. Determine the charge on each bead, if $m=0.001\text{kg}$ and $R=0.001\text{m}$

• A. $2\times {10}^{-10}\text{C}$
• B. $4\times {10}^{-10}\text{C}$
• C. $6\times {10}^{-10}\text{C}$
• D. $8\times {10}^{-10}\text{C}$

Answer 1

The correct option is D $8\times {10}^{-10}\text{C}$

Given:
$m=0.001\text{kg};R=0.001\text{m}$

A repulsive coulomb force $F$ is acting on the charges. So,

From Coulomb's law,

$F=\frac{k{q}^2}{R^2}.........\left(1\right)$


Since, both charges are in equilibrium and both experiencing three different forces which is shown in the above diagram.

Dividing Normal into components

$N\sin {30}^{\circ }=F$
$N\cos {30}^{\circ }=mg$
$\Rightarrow \frac{F}{mg}=\tan {30}^{\circ }$

$\Rightarrow F=\frac{mg}{\sqrt{3}}$

$\Rightarrow \frac{k{q}^2}{R^2}=\frac{mg}{\sqrt{3}}$ (from Eq. (1))

$\Rightarrow q=\sqrt{\frac{mg{R}^2}{\sqrt{3}k}}$

putting the value of given data

$\Rightarrow q=\sqrt{\frac{0.001\times 10\times (0.001{)}^2}{\sqrt{3}\times 9\times {10}^9}}$

$\Rightarrow q=8\times {10}^{-10}\text{C}$

Hence, option $\left(d\right)$ is the correct answer.

(OR)
Applying Lami's Theorem on charge, we have,

$\frac{mg}{\sin ({90}^{\circ }+{30}^{\circ })}=\frac{F}{\sin ({90}^{\circ }+{60}^{\circ })}$

$\Rightarrow F=\frac{mg}{\sqrt{3}}$

$\Rightarrow \frac{k{q}^2}{R^2}=\frac{mg}{\sqrt{3}}$ (from Eq. (1))

$\Rightarrow q=\sqrt{\frac{mg{R}^2}{\sqrt{3}k}}$

putting the value of given data

$\Rightarrow q=\sqrt{\frac{0.001\times 10\times (0.001{)}^2}{\sqrt{3}\times 9\times {10}^9}}$

$\Rightarrow q=8\times {10}^{-10}\text{C}$

Hence, option $\left(d\right)$ is the correct answer.