Two points $P$ and $Q$ are maintained at the potentials of $10\, V$ and $-4\,V$, respectively. The work done in moving $100$ electrons from $P$ and $Q$ is

When three electric dipoles are near each other, they each experience the electric field of the other two, and the three dipole system has a certain potential energy. Figure below shows three arrangements $(1)$ , $(2)$ and $(3)$ in which three electric dipoles are side by side. All three dipoles have the same magnitude of electric dipole moment, and the spacings between adjacent dipoles are identical. If $U_1$ , $U_2$ and $U_3$ are potential energies of the arrangements $(1)$ , $(2)$ and $(3)$ respectively then

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)

$(a)$ Calculate the potential at a point $P$ due to a charge of $4 \times 10^{-7}\; C$ located $9 \;cm$ away.

$(b)$ Hence obtain the work done in bringing a charge of $2 \times 10^{-9} \;C$ from infinity to the point $P$. Does the answer depend on the path along which the charge is brought?

A circle of radius $R$ is drawn with charge $+ q$ at the centre. A charge $q_0$ is brought from point $B$ to $C$, then work done is

Kinetic energy of an electron accelerated in a potential difference of $100\, V$ is