Question

A charged particle $q$ is shot from a large distance towards another charged particle $Q$ which is fixed, with a speed $v$. lt approaches $Q$ up to a closest distance $r$ and then returns. If $q$ were given a speed $2v$, the distance of approach would be

• A. $r$
• B. $2r$
• C. $r/2$
• D. $r/4$

Answer 1

The correct option is D $r/4$

For first case, change in potential energy at closest diatance is $kQq/r$.

kinetic energy is $m{v}^2/2$

By conservation of energy, $kQq/r=m{v}^2/2\Rightarrow r=2kQq/m{v}^2-\left(1\right)$

In second case, when charge is launched with velocity $2v$, let closest distance be $r_1$

similar to above case,

By conservation of energy, $kQq/r_1=m(2v{)}^2/2\Rightarrow r_1=2kQq/4m{v}^2$

From equation $\left(1\right)$, $r_1=r/4$