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Question: A charged particle of mass \[{m_1}\] and charge \[{q_1}\] is revolving in a circle of radius \[r\]. ...

A charged particle of mass m1{m_1} and charge q1{q_1} is revolving in a circle of radius rr. Another charged particle of charge q2{q_2} and mass m2{m_2} is situated at the centre of the circle. If the velocity and time period of the revolving particle be vv and TT respectively, then:
A. v=q1q2r4πε0m1v = \sqrt {\dfrac{{{q_1}{q_2}r}}{{4\pi {\varepsilon _0}{m_1}}}}
B. v=1m1q1q24πε0rv = \dfrac{1}{{{m_1}}}\sqrt {\dfrac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}r}}}
C. T=16π3ε0m12r3q1q2T = \sqrt {\dfrac{{16{\pi ^3}{\varepsilon _0}m_1^2{r^3}}}{{{q_1}{q_2}}}}
D. T=16π3ε0m2r3q1q2T = \sqrt {\dfrac{{16{\pi ^3}{\varepsilon _0}{m_2}{r^3}}}{{{q_1}{q_2}}}}
E. None of the above

Explanation

Solution

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 FC{F_C} acting on an object in circular motion is
FC=mv2R{F_C} = \dfrac{{m{v^2}}}{R} …… (1)
Here, mm is the mass of the object, vv is the velocity of the object and RR is the radius of the circular path.
The electrostatic force FF between the two charges q1{q_1} and q2{q_2} is
F=14πε0q1q2r2F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}} …… (2)
Here, ε0{\varepsilon _0} is the permittivity of free space and rr is the distance between the two charges.
The relation between the linear velocity vv and angular velocity ω\omega is
v=Rωv = R\omega …… (3)
Here, RR is the radius of the circular path.
The angular velocity ω\omega of an object is
ω=2πT\omega = \dfrac{{2\pi }}{T} …… (4)
Here, TT is the time period of the object.

Complete step by step answer:
We have given that a charged particle of mass m1{m_1} and charge q1{q_1} is revolving in a circle of radius rr.Another charged particle of charge q2{q_2} and mass m2{m_2} is situated at the centre of the circle.The velocity and time period of the revolving charged particle are vv and TT respectively.The centripetal force acting on the charged particle q1{q_1} is given by
FC=m1v2r{F_C} = \dfrac{{{m_1}{v^2}}}{r}
The electrostatic force of attraction between the two charged particles is
F=14πε0q1q2r2F = \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.
FC=F{F_C} = F

Substitute m1v2r\dfrac{{{m_1}{v^2}}}{r} for FC{F_C} and 14πε0q1q2r2\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}} for FF in the above equation.
m1v2r=14πε0q1q2r2\dfrac{{{m_1}{v^2}}}{r} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}
v=14πε0q1q2m1r\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 rωr\omega for vv in the above equation.
rω=14πε0q1q2m1r\Rightarrow r\omega = \sqrt {\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{m_1}r}}}

Substitute 2πT\dfrac{{2\pi }}{T} for ω\omega in the above equation.
r2πT=14πε0q1q2m1r\Rightarrow r\dfrac{{2\pi }}{T} = \sqrt {\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{m_1}r}}}
T=2πr14πε0q1q2m1r\Rightarrow T = \dfrac{{2\pi r}}{{\sqrt {\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{m_1}r}}} }}
T=4π2r214πε0q1q2m1r\Rightarrow T = \dfrac{{\sqrt {4{\pi ^2}{r^2}} }}{{\sqrt {\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{m_1}r}}} }}
T=16π3ε0m1r3q1q2\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.