Three point charges shown in the figure lie along a straight line. The energy required to exchange the position of the central charge with one of the negative charges is

Two equal charges of magnitude $Q$ each are placed at a distance $d$ apart. Their electrostatic energy is $E$. $A$ third charge $-Q / 2$ is brought midway between these two charges. The electrostatic energy of the system is now

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

Two charges $+3.2 \times 10^{-19} \, C$ and $-3.2 \times 10^{-19} \, C$ kept $2.4 \, \mathring{A}$ apart form a dipole. If it is kept in a uniform electric field of intensity $4 \times 10^5 \, V/m$,what will be its electrical potential energy in stable equilibrium?

An electric dipole of moment $\overrightarrow{p}$ is lying along a uniform electric field $\overrightarrow{E}$. The work done in rotating the dipole through $90^{\circ}$ is $\left[\sin 0^{\circ}=\cos 90^{\circ}=0, \cos 0^{\circ}=\sin 90^{\circ}=1\right]$

In the first configuration $(1)$ as shown in the figure,four identical charges $(q_0)$ are kept at the corners $A, B, C$ and $D$ of a square of side length $a$. In the second configuration $(2)$,the same charges are shifted to the midpoints $G, E, H$ and $F$ of the sides of the square. If $K = \frac{1}{4 \pi \varepsilon_0}$,the difference between the potential energies of configuration $(2)$ and $(1)$ is given by: