A proton has a mass $1.67 \times 10^{-27} \,kg$ and charge $+1.6 \times 10^{-19} \,C$. If the proton is accelerated through a potential difference of million volts, then the kinetic energy is ......... $J$
$1.6 \times 10^{-15}$
$1.6 \times 10^{-13}$
$1.6 \times 10^{-21}$
$3.2 \times 10^{-13}$
A point charge $q$ of mass $m$ is suspended vertically by a string of length $l$. A point dipole of dipole moment $\overrightarrow{ p }$ is now brought towards $q$ from infinity so that the charge moves away. The final equilibrium position of the system including the direction of the dipole, the angles and distances is shown in the figure below. If the work done in bringing the dipole to this position is $N \times( mgh )$, where $g$ is the acceleration due to gravity, then the value of $N$ is. . . . . . (Note that for three coplanar forces keeping a point mass in equilibrium, $\frac{F}{\sin \theta}$ is the same for all forces, where $F$ is any one of the forces and $\theta$ is the angle between the other two forces)
Figure shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on $r$ for $r / a>>1,$ and contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge).
The diagram shows three infinitely long uniform line charges placed on the $X, Y $ and $Z$ axis. The work done in moving a unit positive charge from $(1, 1, 1) $ to $(0, 1, 1) $ is equal to
Three charges $Q,\, + q$ and $ + q$ are placed at the vertices of a right-angled isosceles triangle as shown. The net electrostatic energy of the configuration is zero if $Q$ is equal to
Two positive charges of magnitude $q$ are placed at the ends of a side $1$ of a square of side $2a$. Two negative charges of the same magnitude are kept at the other corners. Starting from rest, if a charge $Q$, moves from the middle of side $1$ to the centre of square, its kinetic energy at the centre of square is