Question

A thin disc of radius $R=50$ cm with a circular hole (of radius $r=0.1R$) at its centre is charged to a uniform positive surface charge density $10\mu C{m}^{-2}$ and is placed in y-z plane with the origin at the centre of the disc as shown in the figure. A particle of mass $1g$ with charge $-10mC$ which is free to move only along the x-axis, is placed at a distance $x=0.01R$ and released at $t=0s$. Then:

• A. at time $t=300\mu s,$the speed of the electron will be reduced to zero
• B. at the origin O, kinetic energy of the electron is zero
• C. When electron crosses the origin, its kinetic energy reaches the maximum value, equal to 1.28 J
• D. the electron executes a periodic, non-harmonic motion along x-axis

Answer 1

Answer: (A), (C)

at time $t=300\mu s,$the speed of the electron will be reduced to zero
When electron crosses the origin, its kinetic energy reaches the maximum value, equal to 1.28 J

$E_1=\frac{\sigma }{2{\epsilon }_0}[1-\frac{x}{\sqrt{x^2+R^2}}],along+x-axis$

$E_2=-\frac{\sigma }{2{\epsilon }_0}[1-\frac{x}{\sqrt{x^2+r^2}}],along-x-axis$

$\therefore E=-\frac{\sigma }{2{\epsilon }_0}[1-1+\frac{x}{\sqrt{x^2+r^2}}-\frac{x}{\sqrt{x^2+R^2}}]=\frac{\sigma x}{2{\epsilon }_0}[\frac{1}{\sqrt{x^2+r^2}}-\frac{1}{\sqrt{x^2+R^2}}]$

At $x<<r$, $E=\frac{\sigma x}{2{\epsilon }_0}(\frac{1}{r}-\frac{1}{R})=\frac{\sigma .9x}{2{\epsilon }_{0}R}=\frac{9\sigma }{2{\epsilon }_{0}R}x$

Let 'm' be the ,mass of the particle of charge -q

Then, $m\frac{d^{2}x}{d{t}^2}=-\frac{9q\sigma }{2{\epsilon }_{0}R}.x$

or$\frac{d^{2}x}{d{t}^2}=\frac{-9q\sigma }{2{\epsilon }_{0}mR}.x\Rightarrow SHM$

$\omega =\sqrt{\frac{9e\sigma }{2{\epsilon }_{0}m}.\frac{1}{R}}=\frac{2\pi }{T}$

$\therefore T=\sqrt{\frac{8{\pi }^2{\epsilon }_{0}mR}{9q\sigma }}$

Now,$\frac{1}{4\pi {\epsilon }_0}=9\times {10}^9,m=1\times {10}^{-3}kg,R=0.5m,q=10\times {10}^{-3}C,\sigma =10\times {10}^{-6}c{m}^{-2}$

$T^2=\frac{2\pi \times 1\times {10}^{-3}\times 0.5}{9\times {10}^9\times 9\times 10\times {10}^{-3}\times 10\times {10}^{-6}}=\frac{10\pi }{81}\times {10}^{-6}$

$\therefore T=\frac{1}{9}\sqrt{10\pi }\times {10}^{-3}=\frac{5.6}{9}\times {10}^{-3}$

At $t=\frac{T}{2}$, the particle is at an extreme point on the other side, where its kinetic energy and hence its speed becomes zero.

$\therefore t=\frac{T}{2}=\frac{2.8}{9}\times {10}^{-3}=0.31\times {10}^{-3}s$

$\simeq 300\times {10}^{-6}s=300\mu s$

At O, the kinetic energy of the particle is, maximum and is given by,

$K_{max}=\frac{1}{2}m{\omega }^2{A}^2,A\left(amplitude\right)=0.5cm$

$=\frac{1}{2}\times 1\times {10}^{-3}\times \frac{4{\pi }^2\times 81\times {10}^6}{10\pi }\times 25\times {10}^{-6}$

$=\frac{\pi \times 81}{2}\times {10}^{-2}\simeq 1.28J$