$A$ charged particle $q$ is shot towards another charged particle $Q$ which is fixed,with a speed $v$. It approaches $Q$ up to a closest distance $r$ and then returns. If $q$ were given a speed $2v$,the closest distance of approach would be

$A$ $3 \text{ C}$ charge moves from the point $(0, -2, -5)$ to the point $(5, 1, 2)$ in an electric field expressed as $\vec{E} = 2\hat{i} + 3\hat{j} + 4\hat{k} \text{ N/C}$. The work done in moving the charge is . . . . . . $J$.

An electron is moving with constant velocity along the $x$-axis. If a uniform electric field is applied along the $y$-axis,then its path in the $x-y$ plane will be:

Calculate the work done by the electrostatic force on an electron moving in a circular orbit around the nucleus.

$A$ particle of mass $M$ and charge $q$ is released from rest in a region of uniform electric field of magnitude $E$. After a time $t$, the distance travelled by the charge is $S$ and the kinetic energy attained by the particle is $T$. Then, the ratio $T/S$:

There is a uniform electric field of strength $10^3 \ Vm^{-1}$ along the $Y$-axis. $A$ body of mass $1 \ g$ and charge $10^{-6} \ C$ is projected into the field from the origin along the positive $X$-axis with a velocity of $10 \ ms^{-1}$. Its speed in $ms^{-1}$ after $10 \ s$ is (Neglect gravitation).