Question

A particle of mass m and carrying charge -q1 is moving around a charge +q2 along a circular path of radius r . Find period of revolution of the charge - q1

Answer 1

Hii friend,

# Answer-

T=4πr √[(mr.πε)/q1q2]

# Explaination-

The charge -q is revolving around charge +q thus,

Electrostatic force = (1/4πε)(q1q2/r^2)

Centripetal force = mv^2/r

For a charge in stable orbit

Electrostatic force = Centripetal force

(1/4πε)(q1q2/r^2) = mv^2/r

v = √[q1q2/(mr.4πε)]

Period of revolution of charge -q is

T=2πr/v

T=2πr/√[q1q2/(mr.4πε)]

T=2πr √[(mr.4πε)/q1q2]

T=4πr √[(mr.πε)/q1q2]

Period of revolution is T=4πr √[(mr.πε)/q1q2].

Hope this is useful...

Answer 2

Answer: The period of revolution of the first charge is $T = 4\pi r\sqrt{\dfrac{\pi\epsilon_{0}mr}{q_{1}q_{2}}}$.

Explanation:

Given that,

First charge = $-q_{1}$

Second charge = $+q_{2}$

Radius = r

Mass = m

We know that,

First charge revolving around the second charge

Then,

Electrostatic force is

$F_{E}= \dfrac{1}{4\pi\epsilon_{0}}\dfrac{q_{1}q_{2}}{r^2}$

Centripetal force is

$F_{c}= \dfrac{mv^2}{r}$

When a charge is in stable orbit

Electrostatic force = centripetal force

$\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q_{1}q_{2}}{r^2}= \dfrac{mv^2}{r}$

$v = \sqrt{\dfrac{q_{1}q_{2}}{4\pi\epsilon_{0}mr}}$....(I)

The period of revolution of the first charge

$T = \dfrac{2\pi r}{v}$

Put the value of v from equation (I)

$T = \dfrac{2\pi r}{\sqrt{\dfrac{q_{1}q_{2}}{4\pi\epsilon_{0}mr}}}$

$T = 2\pi r\sqrt{\dfrac{4\pi\epsilon_{0}mr}{q_{1}q_{2}}$

$T = 4\pi r\sqrt{\dfrac{\pi\epsilon_{0}mr}{q_{1}q_{2}}}$

Hence, The period of revolution of the first charge is $T = 4\pi r\sqrt{\dfrac{\pi\epsilon_{0}mr}{q_{1}q_{2}}}$.