Question

Consider two point charges, $\frac{q}{3}$ and $-\frac{2q}{3}$ at point $\text{B}$ and $\text{C}$ respectively as shown in the figure. Take $O$ to be the centre of the circle of radius $R$ and $\angle CAB={60}^{\circ }$. The magnitude of force between the charges is

• A. $\frac{q^2}{54\pi {ϵ}_0{R}^2}$
• B. $\frac{q^2}{27\pi {ϵ}_0{R}^2}$
• C. $\frac{q^2}{25\pi {ϵ}_0{R}^2}$
• D. $\frac{q^2}{60\pi {ϵ}_0{R}^2}$

Answer 1

The correct option is A $\frac{q^2}{54\pi {ϵ}_0{R}^2}$

Given;
$\angle BAC={60}^{\circ };Q_1=\frac{q}{3};Q_2=-\frac{2q}{3}$

Now, $\angle ACB$ will be of ${90}^{\circ }$ because we know that angle subtended by the diameter at any point on the circumference of the circle is ${90}^{\circ }$.

So, In $\Delta ABC$,

$\sin {60}^{\circ }=\frac{BC}{AB}\Rightarrow BC=AB\sin {60}^{\circ }$

$\Rightarrow BC=2R\times \frac{\sqrt{3}}{2}=\sqrt{3}R$

Further, apply Coulomb's law,

$F=\frac{Q_1{Q}_2}{4\pi {ϵ}_0{R}^2}=\frac{\frac{q}{3}\times \frac{2q}{3}}{4\pi {ϵ}_0(\sqrt{3}R{)}^2}$

$\Rightarrow F=\frac{q^2}{54\pi {ϵ}_0{R}^2}$

Hence, option $\left(a\right)$ is correct.