Question

Two circular rings $A$ and $B$, each of radius $a=30cm$ are placed coaxially with their axes horizontal in a uniform electric field $E=10^{5}N/C$ directed vertically upward as shown in figure. Distance between 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 $m=100g$ 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 of $40cm$. (Take $g=10m{s}^{-2}$ ) if it is $vm/s$. Find $v/\sqrt{2}$.

Answer 1

Answer: 6.00

Given
$E=10^{5}N/C$
$a=30cm$
$q_1=10\times 10^{-6}C$
$q_2=-20\times 10^{-6}C$
$m=100gram$
$q=10\times 10^{-6}C$
$g=10m/s^2$
$h=40cm$

Weight $=mg$ is balanced by force $F=qE$
$qE=mg$
Since particle is at centre of $A$, hence no force due to charge on $A$, but ring $B$ will exert force on particle.
So force on the particle is towards $B$.
Let $KE$ of particle $=\frac{1}{2}m{v}^2$ at centre of $B$.
Potential energy $=U_1$ (at A)
$=\frac{q{q}_1}{4\pi {\epsilon }_{0}a}+\frac{q\left(-q_2\right)}{4\pi {\epsilon }_0\sqrt{a^2+h^2}}$
Potential energy $=U_2$ (at B)
$=\frac{q\left(q_1\right)}{4\pi {\epsilon }_{0}a\sqrt{a^2+h^2}}+\frac{q\left(-q_2\right)}{4\pi {\epsilon }_{0}a}$
From energy conservation
$\frac{1}{2}m{v}^2=U_1-U_2$
$v^2=\frac{2\times 3.6}{0.1}=72$
or $v=6\sqrt{2}m/s$