In the figure shown the electric potential energy of the system is: ( $q$ is at the centre of the conducting neutral spherical shell of inner radius $a$ and outer radius $b$ )
$0$
$\frac{{{\text{k}}{{\text{q}}^2}}}{{{\text{2b}}}}$
$\frac{{{\text{k}}{{\text{q}}^2}}}{{{\text{2b}}}} - \frac{{{\text{k}}{{\text{q}}^2}}}{{{\text{2a}}}}$
$\frac{{{\text{k}}{{\text{q}}^2}}}{{{\text{2a}}}} - \frac{{{\text{k}}{{\text{q}}^2}}}{{{\text{2b}}}}$
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
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)
In Millikan's experiment, an oil drop having charge $q$ gets stationary on applying a potential difference $V$ in between two plates separated by a distance $d$. The weight of the drop is
When a proton is accelerated through $1\,V$, then its kinetic energy will be.....$eV$
On rotating a point charge having a charge $q$ around a charge $Q$ in a circle of radius $r$. The work done will be