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A Positive Charge Q is Distributed Uniformly Over a Circular Ring of Radius R. a Particle of Mass M, and a Negative Charge Q, is Placed on Its Axis at a Distance X from the Centre. - Physics

Short Note

A positive charge Q is distributed uniformly over a circular ring of radius R. A particle of mass m, and a negative charge q, is placed on its axis at a  distance x from the centre. Find the force on the particle. Assuming x << R, find the time period of oscillation of the particle if it is released from there . 

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Solution

Consider an element of angular width \[d\theta\]  at a distance r from the charge q on the circular ring, as shown in the figure. 

So, charge on the element,
\[dq = \frac{Q}{2\pi R}Rd\theta = \frac{Q}{2\pi}d\theta\]
Electric field due to this charged element,

\[dE = \frac{dq}{4\pi \epsilon_0}\frac{1}{r^2}\] 

\[           = \frac{\frac{Q}{2\pi}d\theta}{4\pi \epsilon_0}\frac{1}{R^2 + x^2}\]

By the symmetry, the E sinθ component of all such elements on the ring will vanish.
So, net electric field,

\[d E_{net}  = dE\cos\theta = \frac{Qd\theta}{8 \pi^2 \epsilon_0 \left( R^2 + x^2 \right)^{3/2}}\]

Total force on the charged particle,

\[F = \int qd E_{net} \] 

\[       = \frac{qQ}{8 \pi^2 \epsilon_0}\frac{x}{\left( R^2 + x^2 \right)^{3/2}} \int_0^{2\pi} d\theta\] 

\[         = \frac{xQq}{4\pi \epsilon_0 \left( R^2 + x^2 \right)^{3/2}}\]

According to the question,

\[x <  < R\] 

\[F = \frac{Qq  x}{4\pi \epsilon_0 R^3}\] 

Comparing this with the condition of simple harmonic motion, we get

\[F = m \omega^2 x\] 

\[ \Rightarrow m \omega^2  = \frac{Qq}{4\pi \epsilon_0 R^3}\] 

\[ \Rightarrow m \left( \frac{2\pi}{T} \right)^2  = \frac{Qq}{4\pi \epsilon_0 R^3}\] 

\[ \Rightarrow T =  \left[ \frac{16 \pi^3 \epsilon_0 m R^3}{Qq} \right]^{1/2}\]

 
Concept: Electric Charges
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APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 7 Electric Field and Potential
Q 39 | Page 122
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