Two charges $7 \ \mu C$ and $-4 \ \mu C$ are placed at $(-7 \ cm, 0, 0)$ and $(7 \ cm, 0, 0)$ respectively. Given $\varepsilon_0 = 8.85 \times 10^{-12} \ C^2 \ N^{-1} \ m^{-2}$,the electrostatic potential energy of the charge configuration is: (in $J$)

An electric dipole with dipole moment $\vec p$ is placed in a uniform electric field $\vec E$. The work done in rotating the dipole by $90^\circ$ from its initial equilibrium position is:

The electric potential at a point on the axis of an electric dipole depends on the distance $r$ of the point from the dipole as

Two charges $-q$ each are separated by a distance $2d$. $A$ third charge $+q$ is kept at the midpoint $O$. Find the potential energy of $+q$ as a function of a small distance $x$ from $O$ due to the $-q$ charges. Sketch the $P.E.$ versus $x$ graph and convince yourself that the charge at $O$ is in an unstable equilibrium.

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?

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