Question

Three equal point charges with charge $+q$ each are moving along a circle of radius $R$ and a fixed point charge $-2q$ is placed at the centre of the circle. If $+q$ charges are revolving at constant speed around $-2q$ as shown, then the speed of charges are
(Assume that charges $(+q)$ form vertices of equilateral triangle)

• A. $\frac{k{q}^2}{\sqrt{3}Rm}$
• B. $\sqrt{\frac{k{q}^2}{Rm}(2-\frac{1}{\sqrt{3}})}$
• C. $\sqrt{\frac{k{q}^2}{Rm}(\sqrt{2}+1)}$
• D. $\frac{k{q}^2}{2Rm}$

Answer 1

The correct option is B $\sqrt{\frac{k{q}^2}{Rm}(2-\frac{1}{\sqrt{3}})}$

We can apply coulomb's law in this case, as relative distance between charges remains same during motion .

From the symmetry of the figure, we can see that triangle $ABC$ is an equilateral triangle. Thus, each angle is ${60}^{\circ }$.
Each side of $\Delta ABC$ is, $(a=AB=BC=CA)$
$\Rightarrow a=BC=R\cos {30}^{\circ }+R\cos {30}^{\circ }$
$\therefore BC=2R\cos {30}^{\circ }=\sqrt{3}R$
$\Rightarrow a=\sqrt{3}R$
Again from symmetry, the repulsive force experienced by charge at $A$ from charges at corner $B$ and $C$ will be same.

The force experienced by charge at $A$ due to charge at $B$ or $C$ is
$F_1=\frac{k\left(q\right)\left(q\right)}{(\sqrt{3}R{)}^2}=\frac{k{q}^2}{3R^2}$

The force experienced by charge $A$ due to charge at center is
$F_2=\frac{kq\left(2q\right)}{R^2}=\frac{2k{q}^2}{R^2}$

$\Rightarrow$ Net force on charge at $A$ is toward centre.
$F_c=F_2-2F_1\cos {30}^{\circ }$ (Horizontal components are cancelled)

Let charges are revolving at the speed of $v$.
Then
$F_{\text{centripetal}}=\frac{m{v}^2}{R}$
$\Rightarrow F_c=\frac{m{v}^2}{R}$
$\Rightarrow \frac{2k{q}^2}{R^2}-2.\frac{k{q}^2}{3R^2}.\frac{\sqrt{3}}{2}=\frac{m{v}^2}{R}$
$\Rightarrow \frac{2\sqrt{3}k{q}^2-k{q}^2}{\sqrt{3}{R}^2}=\frac{m{v}^2}{R}$
$\Rightarrow \frac{k{q}^2}{R}[2-\frac{1}{\sqrt{3}}]=m{v}^2$
$\therefore v=\sqrt{\frac{k{q}^2}{mR}[2-\frac{1}{\sqrt{3}}]}$

$\begin{array}{c}\text{Why this question ?}\\ \text{Tip: It challenges you to simplify the problem by}\\ \text{using symmetry of circular path and then apply equation}\\ \text{of dynamics towards the centre of circle}\end{array}$