Question

Charge $​ { q }_{ 2 }$ of mass m revolves around a stationary charge $​ { q }_{ 1 }$ ​ in a circular orbit of radius r. The orbital periodic time of $​ { q }_{ 2 }$ would be.
$A. { \left( \dfrac { 4{ \pi }^{ 2 }m{ r }^{ 3 } }{ k{ q }_{ 1 }{ q }_{ 2 } } \right) }^{ \dfrac { 1 }{ 2 } }$
$B. { \left( \dfrac { k{ q }_{ 1 }{ q }_{ 2 } }{ 4{ \pi }^{ 2 }m{ r }^{ 3 } } \right) }^{ \dfrac { 1 }{ 2 } }$
$C. { \left( \dfrac { 4{ \pi }^{ 2 }m{ r }^{ 4 } }{ k{ q }_{ 1 }{ q }_{ 2 } } \right) }^{ \dfrac { 1 }{ 2 } }$
$D. { \left( \dfrac { 4{ \pi }^{ 2 }m{ r }^{ 2 } }{ k{ q }_{ 1 }{ q }_{ 2 } } \right) }^{ \dfrac { 1 }{ 2 } }$

Answer 1

Hint: Force acting on the particle can be balanced by equating Electrostatic force and centripetal force on the charge. By equating you get the value for v. Use the relation between velocity, displacement and time. Substitute the value of calculated velocity and displacement and find time.

Formula used:
${ F }_{ e }=\dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } \dfrac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } }$
${ F }_{ c }=\dfrac { m{ v }^{ 2 } }{ r }$

Complete step by step answer:
Let the charge be moving with velocity ‘v’.
The total path for revolution (x) is $2\pi r$. …(1)
Electrostatic force is given by,
${ F }_{ e }=\dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } \dfrac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } }$ …(2)
 Centripetal Force is given by,
${ F }_{ c }=\dfrac { m{ v }^{ 2 } }{ r }$ …(3)
 For stable orbit, these forces should be equal.
$\therefore { F }_{ e }={ F }_{ c }$
 By equating equation.(1) and equation.(2) we get,
$\dfrac { 1 }{ 4\pi \epsilon _{ 0 } } \dfrac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } =\dfrac { m{ v }^{ 2 } }{ r }$
$\therefore v = { \left( \dfrac { { q }_{ 1 }{ q }_{ 2 } }{ 4m\pi { \epsilon }_{ 0 }r } \right) }^{ \dfrac { 1 }{ 2 } }$ …(4)
Now, we know $v= \dfrac { x }{ t }$ …(5)
By substituting equation.(1) and equation.(2) in equation.(5) we get,
$ { \left( \dfrac { { q }_{ 1 }{ q }_{ 2 } }{ 4m\pi { \epsilon }_{ 0 }r } \right) }^{ \dfrac { 1 }{ 2 } }= \dfrac { 2\pi r }{ t }$
$\therefore t= { \left( \dfrac { 16{ \pi }^{ 3 }{ \epsilon }_{ 0 }m{ r }^{ 3 } }{ { q }_{ 1 }{ q }_{ 2 } } \right) }^{ \dfrac { 1 }{ 2 } }$
But $\dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } = k$
$\therefore t= { \left( \dfrac { 4{ \pi }^{ 2 }m{ r }^{ 3 } }{ { kq }_{ 1 }{ q }_{ 2 } } \right) }^{ \dfrac { 1 }{ 2 } }$
Therefore, the orbital periodic time of $​ { q }_{ 2 }$ would be $ { \left( \dfrac { 4{ \pi }^{ 2 }m{ r }^{ 3 } }{ k{ q }_{ 1 }{ q }_{ 2 } } \right) }^{ \dfrac { 1 }{ 2 } }$

Hence, the correct answer is option A i.e. ${ \left( \dfrac { 4{ \pi }^{ 2 }m{ r }^{ 3 } }{ k{ q }_{ 1 }{ q }_{ 2 } } \right) }^{ \dfrac { 1 }{ 2 } }$.

Note:
There is an alternate method to solve this problem. In alternate method, you can take centripetal force as, ${ F }_{ c }= mr{ \omega }^{ 2 }$
But, $\omega =\dfrac { 2\pi }{ T }$
where, T: Time Period
 $\therefore { F }_{ c }= \dfrac { 4mr{ \pi }^{ 2 } }{ { T }^{ 2 } }$
 Now you can equate the Electrostatic Force and Centripetal Force and calculate T.
Thus, you can calculate orbital time period using this method.