Two charges ${q_1}$ and ${q_2}$ are placed $30\,cm$ apart,as shown in the figure. $A$ third charge ${q_3}$ is moved along the arc of a circle of radius $40\,cm$ from $C$ to $D$. The change in the potential energy of the system is $\frac{{{q_3}}}{{4\pi {\varepsilon _0}}}k$,where $k$ is:

The distance between $H^+$ and $Cl^-$ ions in an $HCl$ molecule is $1.28 \, \mathring{A}$. What will be the electric potential due to this dipole at a distance of $12 \, \mathring{A}$ on the axis of the dipole (in $, V$)?

Two equal charges $q$ of opposite sign separated by a distance $2a$ constitute an electric dipole of dipole moment $p$. If $P$ is a point at a distance $r$ from the centre of the dipole and the line joining the centre of the dipole to this point makes an angle $\theta$ with the axis of the dipole,then the potential at $P$ is given by $(r \gg 2a)$ (where $p = 2qa$).

Two charges $7 \ \mu C$ and $-4 \ \mu C$ are placed at $(-7 \ cm, 0, 0)$ and $(7 \ cm, 0, 0)$ respectively. Given $\varepsilon_0 = 8.85 \times 10^{-12} \ C^2 \ N^{-1} \ m^{-2}$,the electrostatic potential energy of the charge configuration is: (in $J$)

$A$ proton is moving away from an electron. Find the change in the potential energy of the system.

The potential energy of a system of two equal negative point charges of $2 \mu C$ each,held $1 \ m$ apart in air,is . . . . . . . $\left( k = 9 \times 10^9 \text{ SI units} \right)$