Q.Two point charges q and $-q$ are located at points $(0, 0, -a)$ and $(0, 0, a)$ respectively.

Dipole (moment $p=2aq$): $V(0,0,z)=\dfrac{-2aq}{4\pi\varepsilon_0(z^2-a^2)}$, $V(x,y,0)=0$; work between equatorial points $=0$ (path-independent, conservative field); unstable-equilibrium PE $=+pE$. (OR: capacitor energy loss $=\tfrac12\tfrac{C_1C_2}{C_1+C_2}(V_1-V_2)^2$.)

Setup. $+q$ at $(0,0,-a)$ and $-q$ at $(0,0,a)$ form a dipole of moment $p=q(2a)$ pointing along $-z$.

(a) Potentials. With $k=\dfrac{1}{4\pi\varepsilon_0}$:

• On the axis $(0,0,z)$ (take $z>a$): distances $z+a$ and $z-a$, so $V=kq\left(\frac{1}{z+a}-\frac{1}{z-a}\right)=\frac{kq(-2a)}{z^2-a^2}=\frac{-2aq}{4\pi\varepsilon_0(z^2-a^2)}.$
• On the plane $(x,y,0)$: each charge is equidistant, $r=\sqrt{x^2+y^2+a^2}$, so $V=\dfrac{kq}{r}-\dfrac{kq}{r}=0.$

(b) Work $(5,0,0)\to(-7,0,0)$. Both points lie on the $z=0$ plane, where $V=0$. Work $=q_0(V_f-V_i)=q_0(0-0)=0.$

(c) Any other path. The electrostatic force is conservative, so work depends only on the endpoints, not the path. Hence the work is still zero.

(d) Unstable equilibrium in a uniform field $\vec E$. Potential energy of a dipole: $U=-\vec p\cdot\vec E=-pE\cos\theta$. Unstable equilibrium is when $\vec p$ is antiparallel to $\vec E$ ($\theta=180^\circ$):

$U_{\text{unstable}}=+pE=+2aqE.$

OR-alternative (capacitors). …