Two identical thin rings each of radius $R$ meters are coaxially placed at a distance $R$ meters apart. If $Q_1$ coulomb and $Q_2$ coulomb are respectively the charges uniformly spread on the two rings, the work done in moving a charge $q$ from the centre of one ring to that of other is
Zero
$\frac{{q({Q_1} - {Q_2})(\sqrt 2 - 1)}}{{\sqrt 2 .4\pi {\varepsilon _0}R}}$
$\frac{{q\sqrt 2 ({Q_1} + {Q_2})}}{{4\pi {\varepsilon _0}R}}$
$\frac{{q({Q_1} + {Q_2})(\sqrt 2 + 1)}}{{\sqrt 2 .4\pi {\varepsilon _0}R}}$
The work done to take an electron from rest where potential is $-60\, V$ to another point where potential is $-20\, V$ is given by.....$eV$
The diagram shows a small bead of mass $m$ carrying charge $q$. The bead can freely move on the smooth fixed ring placed on a smooth horizontal plane. In the same plane a charge $+Q$ has also been fixed as shown. The potential atthe point $P$ due to $+Q$ is $V$. The velocity with which the bead should projected from the point $P$ so that it can complete a circle should be greater than
A bullet of mass $2\, gm$ is having a charge of $2\,\mu C$. Through what potential difference must it be accelerated, starting from rest, to acquire a speed of $10\,m/s$
Two electrons are moving towards each other, each with a velocity of $10^6 \,m / s$. What will be closest distance of approach between them is ......... $m$
Six charges $+ q ,- q ,+ q ,- q ,+ q$ and $- q$ are fixed at the corners of a hexagon of side $d$ as shown in the figure. The work done in bringing a charge $q _0$ to the centre of the hexagon from infinity is :$\left(\varepsilon_0-\right.$ permittivity of free space)