Question

Four identical charged particles having same mass are located at four corners of a regular tetrahedron. If only one of the particles is released, with the remaining tetrahedron fixed, it acquires a final speed of $V_o.$ If all the charges are released together, their final speed will be $\frac{V_o}{n}$, then n is

Answer 1

Case 1 : When only one of the particles is released, with the remaining tetrahedron fixed:

According to the conservation of energy,

the total electrostatic potential energy of only one mass = the total kinetic energy of only one mass which is released.
$3\times \frac{K{q}^2}{l}=\frac{1}{2}m{V}_o^2$ ---(1)
Case 2 : When all the charges are released together:

the total electrostatic potential energy of the system = the total kinetic energy of the system
$6\times \frac{K{q}^2}{l}=\frac{4}{2}m{V}^2$ --(2)
From (1) and (2), we get
$V=\frac{V_o}{\sqrt{2}}$

Therefore, the value of $n$ is $\frac{1}{\sqrt{2}}$.