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Grade 12Electrostatics

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

vinayak singh

8 Years agoGrade 12

2 Answers

Eshan

8 Years ago

We are assuming that the charge $+q_2$ is fixed,

The centripetal acceleration of the charge $-q_1$ will be provided by the coulombic attraction by the other charge.
$\implies \dfrac{mv^2}{r}=\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1q_2}{r^2}$
$\implies v=\sqrt{\dfrac{q_1q_2}{4\pi m\epsilon_0r}}$
Hence the time period of revolution= $\dfrac{2\pi r}{v}=\sqrt{\dfrac{16\pi^3 m\epsilon_0 r^3}{q_1q_2}}$

Kushagra Madhukar

5 Years ago

Dear student,

Please find the answer to your question.

The coulombic force will be the source of centripetal force, hence both are equal.

On solving this we get, Period of revolution, T = √16π³mε₀r³/q₁q₂

Thanks and regards,

Kushagra

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