The charges $+q$,$+q$,and $Q$ are placed at the vertices of an equilateral triangle of side $a$. If 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?

Charges $Q$,$+q$,and $+q$ are placed at the vertices of an isosceles right-angled triangle as shown in the figure. If the total electrostatic potential energy of the system is zero,then $Q = $ . . . . . . .

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

An electric dipole is formed by two charges $+q$ and $-q$ located in the $xy$-plane at $(0, 2) \text{ mm}$ and $(0, -2) \text{ mm}$,respectively,as shown in the figure. The electric potential at point $P(100, 100) \text{ mm}$ due to the dipole is $V_0$. The charges $+q$ and $-q$ are then moved to the points $(-1, 2) \text{ mm}$ and $(1, -2) \text{ mm}$,respectively. What is the value of the electric potential at $P$ due to the new dipole?

If $OP = 1 \, cm$ and $OS = 2 \, cm$,calculate the work done by the electric field in shifting a point charge $q = \frac{4\sqrt{2}}{27} \, \mu C$ from point $P$ to $S$ in the given figure. The dipole moment is $p = 2 \times 10^{-6} \, C \cdot m$.