Question

Two circular rings $A$ and $B$, each of radius $a=30 \,cm$ 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 $h=40 \,cm$. 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 $m =100 \,g$ 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 $40\, cm$. (Take $g=10 \, ms ^{-2}$ ) if it is $v m / s$. Find $v / \sqrt{2}$.

Answer 1

Given
$ E =10^{5} \,N / C $
$ a =30\, cm $
$q _{1} =10 \times 10^{-6} C $
$ q _{2} =-20 \times 10^{-6} C $
$ m =100\, gram $
$ q =10 \times 10^{-6} C $
$ g =10\, m / s ^{2} $
$h =40\, cm $

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 $=\frac{1}{2} mv ^{2}$ at centre of $B$.
Potential energy $=U_{1}$ (at A)
$=\frac{q q_{1}}{4 \pi \varepsilon_{0} a}+\frac{q\left(-q_{2}\right)}{4 \pi \varepsilon_{0} \sqrt{a^{2}+h^{2}}}$
Potential energy $=U_{2}$ (at B)
$=\frac{q\left(q_{1}\right)}{4 \pi \varepsilon_{0} a \sqrt{a^{2}+h^{2}}}+\frac{q\left(-q_{2}\right)}{4 \pi \varepsilon_{0} a}$
From energy conservation
$\frac{1}{2} mv ^{2} = U _{1}- U _{2} $
$v ^{2} =\frac{2 \times 3.6}{0.1}=72$
or $ v=6 \sqrt{2} \,m / s$