Question

A positive charge q of mass m is projected from a very large distance towards a fixed non conducting sphere having charge q uniformly distributed inside it if the charge just grazes the sphere then speed of projection

Answer 1

Given:

A positive charge q of mass m is projected from a very large distance towards a fixed non conducting sphere having charge q uniformly distributed inside it. The charge just grazes the sphere.

To find:

Initial speed of projection.

Calculation:

Let initial velocity of projection be u.

Velocity during grazing be v , such that v= u/2 (according to conservation of angular momentum).

Applying Conservation of Mechanical Energy:

$\dfrac{1}{2} m {u}^{2} + \dfrac{k {q}^{2} }{ \infty} = \dfrac{1}{2} m {v}^{2} + \dfrac{k {q}^{2} }{r}$

$= > \dfrac{1}{2} m {u}^{2} + 0 = \dfrac{1}{2} m { (\dfrac{u}{2} )}^{2} + \dfrac{k {q}^{2} }{r}$

$= > \dfrac{1}{2} m {u}^{2} + 0 = \dfrac{1}{8} m { u}^{2} + \dfrac{k {q}^{2} }{r}$

$= > \dfrac{1}{2} m {u}^{2} - \dfrac{1}{8} m { u}^{2} = \dfrac{k {q}^{2} }{r}$

$= > ( \dfrac{4 - 1}{8} )m {u}^{2} = \dfrac{k {q}^{2} }{r}$

$= > \dfrac{3}{8} m {u}^{2} = \dfrac{k {q}^{2} }{r}$

$= > {u}^{2} = \dfrac{8k {q}^{2} }{3mr}$

$= > {u}^{2} = \dfrac{8 {q}^{2} }{3mr \times 4\pi\epsilon_{0}}$

$= > {u}^{2} = \dfrac{2 {q}^{2} }{3mr \pi\epsilon_{0}}$

$= > u = \sqrt{ \dfrac{2 {q}^{2} }{3mr \pi\epsilon_{0}} }$

So, final answer is :

$\boxed{ \red{ \bold{ \large{ u = \sqrt{ \dfrac{2 {q}^{2} }{3mr \pi\epsilon_{0}} }}}}}$