Two particles each of mass $m$ and charge $q$ are separated by distance $r_1$ and the system is left free to move at $t = 0$. At time $t$ both the particles are found to be separated by distance $r_2$. The speed of each particle is
$\frac{{qm}}{{4\pi {\varepsilon _0}{r_1}{r_2}}}$
$\frac{q}{{{r_1}{r_2}\sqrt {(r_2^2 - r_1^2)/4\pi {\varepsilon _0m}} }}$
$\frac{\sqrt 2q}{{{r_1}{r_2}\sqrt {(r_2^2 - r_1^2)/4\pi {\varepsilon _0m}} }}$
none of these
Two equal point charges are fixed at $x = - a$ and $x = + a$ on the $x-$axis. Another point charge $Q$ is placed at the origin. The Change in the electrical potential energy of $Q$, when it is displaced by a small distance $x$ along the $x$-axis, is approximately proportional to
Charge $Q$ is given a displacement $\vec r = a\hat i + b\hat j$ in an electric field $\vec E = E_1\hat i + E_2\hat j$ . The work done is
A solid sphere of radius $R$ carries a charge $(Q+q)$ distributed uniformly over its volume. A very small point like piece of it of mass $m$ gets detached from the bottom of the sphere and falls down vertically under gravity. This piece carries charge $q.$ If it acquires a speed $v$ when it has fallen through a vertical height $y$ (see figure), then :
(assume the remaining portion to be spherical).
A particle of charge $q$ and mass $m$ is subjected to an electric field $E = E _{0}\left(1- ax ^{2}\right)$ in the $x-$direction, where $a$ and $E _{0}$ are constants. Initially the particle was at rest at $x=0 .$ Other than the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is
For equal point charges $Q$ each are placed in the $xy$ plane at $(0, 2), (4, 2), (4, -2)$ and $(0, -2)$. The work required to put a fifth change $Q$ at the origin of the coordinate system will be