Question
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 and charge q1 is revolving in a circle of radius r. Another charged particle of charge q2 and mass m2 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=4πε0m1q1q2r
B. v=m114πε0rq1q2
C. T=q1q216π3ε0m12r3
D. T=q1q216π3ε0m2r3
E. None of the above
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 acting on an object in circular motion is
FC=Rmv2 …… (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 q1 and q2 is
F=4πε01r2q1q2 …… (2)
Here, ε0 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 ω is
v=Rω …… (3)
Here, R is the radius of the circular path.
The angular velocity ω of an object is
ω=T2π …… (4)
Here, T is the time period of the object.
Complete step by step answer:
We have given that a charged particle of mass m1 and charge q1 is revolving in a circle of radius r.Another charged particle of charge q2 and mass m2 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 q1 is given by
FC=rm1v2
The electrostatic force of attraction between the two charged particles is
F=4πε01r2q1q2
The centripetal force of acting on the revolving particle balances the electrostatic force of attraction between the charged particles.
FC=F
Substitute rm1v2 for FC and 4πε01r2q1q2 for F in the above equation.
rm1v2=4πε01r2q1q2
⇒v=4πε01m1rq1q2
This is the expression for velocity of the revolving particle.
Substitute rω for v in the above equation.
⇒rω=4πε01m1rq1q2
Substitute T2π for ω in the above equation.
⇒rT2π=4πε01m1rq1q2
⇒T=4πε01m1rq1q22πr
⇒T=4πε01m1rq1q24π2r2
∴T=q1q216π3ε0m1r3
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.