Physics - Electrostatics Question with Solution | TestHub

Initially, at a very large distance, the potential energy is zero ($U_i=0$) and kinetic energy is $K_i=\frac{1}{2}m{v}_0^2$.

At the minimum distance of approach $r_0$, the particle momentarily stops, so its kinetic energy is zero ($K_{r_0}=0$) and its potential energy is $U_{r_0}=k\frac{qQ}{r_0}$.

Thus, $\frac{1}{2}m{v}_0^2=k\frac{qQ}{r_0}$ (1).

When the particle is at a distance $2r_0$, its potential energy is $U_{2r_0}=k\frac{qQ}{2r_0}$ and its kinetic energy is $K_{2r_0}=\frac{1}{2}m{v}^2$.

By energy conservation,

$\frac{1}{2}m{v}_0^2=k\frac{qQ}{2r_0}+\frac{1}{2}m{v}^2$ ( 2). Substituting $k\frac{qQ}{r_0}=\frac{1}{2}m{v}_0^2$ from (1) into (2),

we get

$\frac{1}{2}m{v}_0^2=\frac{1}{2}(\frac{1}{2}m{v}_0^2)+\frac{1}{2}m{v}^2$.

$\frac{1}{2}m{v}_0^2=\frac{1}{4}m{v}_0^2+\frac{1}{2}m{v}^2$,

which gives $\frac{1}{4}m{v}_0^2=\frac{1}{2}m{v}^2$.

Therefore, $v^2=\frac{1}{2}{v}_0^2$, so $v=\frac{v_0}{\sqrt{2}}$.