Question

Two circular rings A and B each of radius $a=30cm$ are placed co-axially with their horizontal in uniform electric field $E={10}^{5}N/C$ directed vertically upward as shown in the figure. The distance the centres of these rings A and B is $h=40cm$. Ring A has a positive charge $q_1=10\mu C$ while ring B has a negative charge of magnitude $q_2=20\mu C$. A particle of mass $m=100gm$ and carrying a positive charge $q=10\mu C$ is released from rest at the centre of the ring A. Calculate its velocity when it has moved a distance $40cm$.

Answer 1

Weight of the particle $W=mg=\frac{100}{1000}\times 10=1$Newton
Force due to electrostatic field $E=qE=10\times {10}^{-6}\times {10}^6=1$Newton.
Thus the weight of the particle is balanced by the electrostatic force.
The particle experiences net force in the line through the centres of ring A nad B and is accelerated towards the centre of the ring B. At the centre of ring B.
$KE=$Loss of $PE$
$⟹\frac{1}{2}m{v}^2=u_A-u_B=\frac{1}{4\pi {\in }_0}\frac{q{q}_1}{a}+\frac{1}{4\pi {\in }_0}\frac{q(-q_2)}{\sqrt{a^2+h^2}}-\frac{1}{4\pi {\in }_0}\frac{q{q}_1}{\sqrt{a^2+h^2}}-\frac{1}{4\pi {\in }_0}\frac{q(-q_2)}{a}$
$\frac{1}{2}m{v}^2=\frac{q}{4\pi {\in }_0}[q_1(\frac{1}{a}-\frac{1}{\sqrt{a^2+h^2}})+q^2(\frac{1}{a}-\frac{1}{\sqrt{a^2+h^2}})]$
$=\frac{q}{4\pi {\in }_0}\left[\right(q_1+q_2\left)\left(\frac{\sqrt{a^2+h^2}-a}{a\sqrt{a^2+h^2}}\right)\right]$
$⟹v=\sqrt{\frac{2q(q_1+q_2)}{4\pi {\in }_{0}m}\left[\frac{\sqrt{a^2+h^2}-a}{a\sqrt{a^2+h^2}}\right]}$
For the given value $v=6\surd 2m/s$.