Question

A charged particle of mass \[{m_1}\] and charge \[{q_1}\] is revolving in a circle of radius \[r\]. Another charged particle of charge \[{q_2}\] and mass \[{m_2}\] is situated at the centre of the circle. If the velocity and time period of the revolving particle be \[v\] and \[T\] respectively, then:
A. \[v = \sqrt {\dfrac{{{q_1}{q_2}r}}{{4\pi {\varepsilon _0}{m_1}}}} \]
B. \[v = \dfrac{1}{{{m_1}}}\sqrt {\dfrac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}r}}} \]
C. \[T = \sqrt {\dfrac{{16{\pi ^3}{\varepsilon _0}m_1^2{r^3}}}{{{q_1}{q_2}}}} \]
D. \[T = \sqrt {\dfrac{{16{\pi ^3}{\varepsilon _0}{m_2}{r^3}}}{{{q_1}{q_2}}}} \]
E. None of the above

Answer 1

Hint:Use the formulae for the centripetal force and electrostatic force of attraction between the two charged particles. Also use the formula for relation between the linear velocity and angular velocity and expression for angular velocity in terms of time period. First calculate the velocity of the revolving particle equating centripetal force and electrostatic force between the particles. Then determine the expression for the time period of the revolving particle.

Formulae used:
The centripetal force \[{F_C}\] acting on an object in circular motion is
\[{F_C} = \dfrac{{m{v^2}}}{R}\] …… (1)
Here, \[m\] is the mass of the object, \[v\] is the velocity of the object and \[R\] is the radius of the circular path.
The electrostatic force \[F\] between the two charges \[{q_1}\] and \[{q_2}\] is
\[F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}\] …… (2)
Here, \[{\varepsilon _0}\] is the permittivity of free space and \[r\] is the distance between the two charges.
The relation between the linear velocity \[v\] and angular velocity \[\omega \] is
\[v = R\omega \] …… (3)
Here, \[R\] is the radius of the circular path.
The angular velocity \[\omega \] of an object is
\[\omega = \dfrac{{2\pi }}{T}\] …… (4)
Here, \[T\] is the time period of the object.

Complete step by step answer:
We have given that a charged particle of mass \[{m_1}\] and charge \[{q_1}\] is revolving in a circle of radius \[r\].Another charged particle of charge \[{q_2}\] and mass \[{m_2}\] is situated at the centre of the circle.The velocity and time period of the revolving charged particle are \[v\] and \[T\] respectively.The centripetal force acting on the charged particle \[{q_1}\] is given by
\[{F_C} = \dfrac{{{m_1}{v^2}}}{r}\]
The electrostatic force of attraction between the two charged particles is
\[F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}\]
The centripetal force of acting on the revolving particle balances the electrostatic force of attraction between the charged particles.
\[{F_C} = F\]

Substitute \[\dfrac{{{m_1}{v^2}}}{r}\] for \[{F_C}\] and \[\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}\] for \[F\] in the above equation.
\[\dfrac{{{m_1}{v^2}}}{r} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}\]
\[ \Rightarrow v = \sqrt {\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{m_1}r}}} \]
This is the expression for velocity of the revolving particle.
Substitute \[r\omega \] for \[v\] in the above equation.
\[ \Rightarrow r\omega = \sqrt {\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{m_1}r}}} \]

Substitute \[\dfrac{{2\pi }}{T}\] for \[\omega \] in the above equation.
\[ \Rightarrow r\dfrac{{2\pi }}{T} = \sqrt {\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{m_1}r}}} \]
\[ \Rightarrow T = \dfrac{{2\pi r}}{{\sqrt {\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{m_1}r}}} }}\]
\[ \Rightarrow T = \dfrac{{\sqrt {4{\pi ^2}{r^2}} }}{{\sqrt {\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{m_1}r}}} }}\]
\[ \therefore T = \sqrt {\dfrac{{16{\pi ^3}{\varepsilon _0}{m_1}{r^3}}}{{{q_1}{q_2}}}} \]
This is the expression for the time period of the revolving particle.

Hence, the correct option is E.

Note:The students should be careful while calculating the time period of the revolving charged particle. The students should not forget to first take the square of the quantities outside the square root and then use the two quantities in the numerator and denominator into the same square root. If these calculations go wrong then the final expression for the time period will be wrong.