Two positive point charges of $10 \ \mu C$ and $12 \ \mu C$ are placed $10 \ cm$ apart in air. The work done to bring them $6 \ cm$ closer is (in $J$)

Two electric dipoles $A$ and $B$ with respective dipole moments $\overrightarrow {{d_A}} = - 4\,qa\,\hat i$ and $\overrightarrow {{d_B}} = 2\,qa\,\hat i$ are placed on the $x-$ axis with a separation $R$,as shown in the figure. The distance from $A$ at which both of them produce the same potential is

An electric dipole of dipole moment $6.0 \times 10^{-6} \, Cm$ is placed in a uniform electric field of $1.5 \times 10^3 \, NC^{-1}$ in such a way that the dipole moment is along the electric field. The work done in rotating the dipole by $180^{\circ}$ in this field will be $......... \, mJ$.

Two point charges $q_1 = 6 \mu\text{C}$ and $q_2 = 4 \mu\text{C}$ are kept at points $A$ and $B$ in air where distance $AB = 10 \text{cm}$. What is the increase in potential energy of the system when $q_2$ is moved towards $q_1$ by $2 \text{cm}$? $(\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{SI units})$

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 $, V$)?

When one electron is moved towards another electron,the electric potential energy of the system: