Concepts Of Physics MCQ Edition [Volume 2]

Exam: Concepts Of Physics MCQ Edition [Volume 2]
Subject: Physics
Chapter: Electric Field and Potential

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Electric Field and Potential

A positive charge Q is distributed uniformly over a circular ring...

A positive charge

Q

is distributed uniformly over a circular ring of radius

R

. A particle having a mass

m

and a negative charge

q

is placed on its axis at a distance

x

from the centre. Determine the force on the particle. Assuming

x \ll R

, find the time period of oscillation of the particle if it is released from this position.

Options

AForce: $\dfrac{1}{4\pi\varepsilon_0} \dfrac{Qqx}{(R^2+x^2)^{3/2}}$ away from the centre; Time period: \left(\dfrac{8 \pi^3 \varepsilon_0 m R^3}{Qq}\right)^{1/2}
BForce: $\dfrac{1}{4\pi\varepsilon_0} \dfrac{Qqx}{(R^2+x^2)^{3/2}}$ away from the centre; Time period: \left(\dfrac{16 \pi^3 \varepsilon_0 m R^3}{Qq}\right)^{1/2}
CForce: $\dfrac{1}{4\pi\varepsilon_0} \dfrac{Qqx}{(R^2+x^2)^{3/2}}$ towards the centre; Time period: \left(\dfrac{8 \pi^3 \varepsilon_0 m R^3}{Qq}\right)^{1/2}
DForce: $\dfrac{1}{4\pi\varepsilon_0} \dfrac{Qqx}{(R^2+x^2)^{3/2}}$ towards the centre; Time period: \left(\dfrac{16 \pi^3 \varepsilon_0 m R^3}{Qq}\right)^{1/2}

Correct answer

D. Force: $\dfrac{1}{4\pi\varepsilon_0} \dfrac{Qqx}{(R^2+x^2)^{3/2}}$ towards the centre; Time period: \left(\dfrac{16 \pi^3 \varepsilon_0 m R^3}{Qq}\right)^{1/2}

Step-by-step solution

The electric field at a distance

x

on the axis of a uniformly charged ring of radius

R and charge

is given by:<br><br>

E = \dfrac{1}{4\pi\varepsilon_0} \dfrac{Qx}{(R^2+x^2)^{3/2}}

<br><br>The force on a particle of negative charge

is:<br><br>

F = qE = \dfrac{1}{4\pi\varepsilon_0} \dfrac{Qqx}{(R^2+x^2)^{3/2}}

<br><br>Since the charge on the particle is negative and the ring is positively charged, the force is attractive and directed towards the centre of the ring.<br><br>For

x \ll R

, we can approximate

(R^2+x^2)^{3/2} \approx R^3

. The restoring force becomes:<br><br>

F = \left(\dfrac{Qq}{4\pi\varepsilon_0 R^3}\right) x

<br><br>This represents simple harmonic motion with force constant

k = \dfrac{Qq}{4\pi\varepsilon_0 R^3}

.<br><br>The time period of oscillation is:<br><br>

T = 2\pi \sqrt{\dfrac{m}{k}} = 2\pi \sqrt{\dfrac{4\pi\varepsilon_0 m R^3}{Qq}} =

\left(\dfrac{16 \pi^3 \varepsilon_0 m R^3}{Qq}\right)^{1/2}

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