Three charges $Q, +q$ and $+q$ are placed at the vertices of a right-angled isosceles triangle with legs of length $a$ as shown in the figure. The net electrostatic potential energy of the configuration is zero. The value of $Q$ is:

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

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 $\left[\sin 0^{\circ}=\cos 90^{\circ}=0, \cos 0^{\circ}=\sin 90^{\circ}=1\right]$

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 $......... \, mJ$.

Three particles,each having a charge of $10 \mu C$,are placed at the corners of an equilateral triangle of side $10 \ cm$. The electrostatic potential energy of the system is (Given $\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \ N \cdot m^{2}/C^{2}$):

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: