The two thin coaxial rings, each of radius $'a'$ and having charges $+{Q}$ and $-{Q}$ respectively are separated by a distance of $'s'.$ The potential difference between the centres of the two rings is :
$\frac{{Q}}{2 \pi \varepsilon_{0}}\left[\frac{1}{{a}}+\frac{1}{\sqrt{{s}^{2}+{a}^{2}}}\right]$
$\frac{{Q}}{4 \pi \varepsilon_{0}}\left[\frac{1}{{a}}+\frac{1}{\sqrt{{s}^{2}+{a}^{2}}}\right]$
$\frac{{Q}}{4 \pi \varepsilon_{0}}\left[\frac{1}{{a}}-\frac{1}{\sqrt{{s}^{2}+{a}^{2}}}\right]$
$\frac{{Q}}{2 \pi \varepsilon_{0}}\left[\frac{1}{{a}}-\frac{1}{\sqrt{{s}^{2}+{a}^{2}}}\right]$
A charge $+q$ is distributed over a thin ring of radius $r$ with line charge density $\lambda=q \sin ^{2} \theta /(\pi r)$. Note that the ring is in the $X Y$ - plane and $\theta$ is the angle made by $r$ with the $X$-axis. The work done by the electric force in displacing a point charge $+ Q$ from the centre of the ring to infinity is
Three concentric spherical shells have radii $a, b$ and $c (a < b < c)$ and have surface charge densities $\sigma ,-\;\sigma $ and $\;\sigma \;$ respectively. If $V_A,V_B$ and $V_C$ denote the potentials of the three shells, then, for $c = a +b,$ we have
The electric potential at the surface of an atomic nucleus $(Z = 50)$ of radius $9.0×{10^{ - 13}}\, cm$ is
Ten charges are placed on the circumference of a circle of radius $R$ with constant angular separation between successive charges. Alternate charges $1,3,5,7,9$ have charge $(+q)$ each, while $2,4,6,8,10$ have charge $(-q)$ each. The potential $V$ and the electric field $E$ at the centre of the circle are respectively
(Take $V =0$ at infinity $)$
Consider a finite insulated, uncharged conductor placed near a finite positively charged conductor. The uncharged body must have a potential