Question

Three identical particles of charges Q and mass m are placed such that they are from an equilateral tringle of side. if they are released simultaneously. their maximum speed attained by any one of the particles will be ( neglect gravity)

• A. $Q\sqrt{\frac{1}{2\pi {\epsilon }_{0}m\ell }}$
• B. $Q\sqrt{\frac{1}{6\pi {\epsilon }_{0}m\ell }}$
• C. Zero
• D. None of these

Answer 1

Answer: (A)

$Q\sqrt{\frac{1}{2\pi {\epsilon }_{0}m\ell }}$

$\begin{array}{c}S\cos {30}^{\circ }=l/2\\ S=\frac{l}{2\cos {30}^{\circ }}\\ S=\frac{l}{\sqrt{3}}\end{array}$

$\begin{array}{c}F=\frac{1}{4\pi {\epsilon }_0}\frac{{\theta }^2}{l^2}\\ {\text{F}}_{\text{net}}\text{along}AB=\sqrt{2t^2+2F^2\cos {60}^{\circ }}\\ =F\sqrt{2+2\times \frac{1}{2}}\\ =F\sqrt{3}\end{array}$

$\begin{array}{c}a=\frac{F_{nct}}{m}=\frac{F\sqrt{3}}{m}\\ v^2=u^2+2as-v\left(1\right)\\ v^2=0^2+2\times \frac{F\sqrt{3}}{m}\times \frac{l}{\sqrt{3}}\\ v^2=2\times \frac{1}{4x{\epsilon }_{0}m}\frac{Q^2}{l^2}\sqrt{3}\frac{l}{\sqrt{3}}\end{array}$

$v=Q\sqrt{\frac{1}{2x{\epsilon }_{0}ml}}$