Three charges $Q, +q$ and $+q$ are placed at the vertices of a right-angled isosceles triangle as shown in the figure. The net electrostatic potential energy of the system is zero. Then $Q$ is equal to:

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

Two charges $-q$ and $+q$ are located at points $(0,0,-a)$ and $(0,0, a)$ respectively.
$(a)$ What is the electrostatic potential at the points $(0,0, z)$ and $(x, y, 0)$?
$(b)$ Obtain the dependence of potential on the distance $r$ of a point from the origin when $r/a > > 1$.
$(c)$ How much work is done in moving a small test charge from the point $(5,0,0)$ to $(-7,0,0)$ along the $x$-axis? Does the answer change if the path of the test charge between the same points is not along the $x$-axis?

Consider the configuration of a system of four charges each of value $+q$. The work done by an external agent in changing the configuration of the system from Figure $(1)$ to Figure $(2)$ is:

Three point charges $+q, +2q$ and $+Q$ are placed at the vertices of an equilateral triangle. If the potential energy of the system of three charges is zero, the value of $Q$ in terms of $q$ is:

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 $J$)