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 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?

Write an equation of electric potential at a point due to an electric dipole.

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 ${q_1}$ 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:

Three charges $Q$,$(-2q)$ and $(-2q)$ are placed at the vertices of an isosceles right-angled triangle as shown in the figure. The net electrostatic potential energy is zero if $Q$ is equal to