In a uniform electric field,the potential is $10 \ V$ at the origin of coordinates,and $8 \ V$ at each of the points $(1, 0, 0), (0, 1, 0)$ and $(0, 0, 1)$. The potential at the point $(1, 1, 1)$ will be....$V$

$A$ uniformly charged solid sphere of radius $R$ has potential $V_0$ (measured with respect to $\infty$) on its surface. For this sphere, the equipotential surfaces with potentials $\frac{3V_0}{2}, \frac{5V_0}{4}, \frac{3V_0}{4}$, and $\frac{V_0}{4}$ have radii $R_1, R_2, R_3$, and $R_4$ respectively. Then:

$1\,eV$ is

Ten electrons are equally spaced and fixed around a circle of radius $R$. Relative to $V = 0$ at infinity,the electrostatic potential $V$ and the electric field $E$ at the centre $C$ are

Two particles with charges $+q$ and $-q$ are separated by a distance of $2d$. What is the electric potential at the midpoint of the line joining them?

$A$ unit positive point charge is taken slowly through an infinitesimally thin tube that is inside a charged dielectric sphere of radius $R$, having uniform positive charge density $\rho$. The initial and final positions of the charge are marked by $A$ and $B$ at distances $2R$ and $3R$ respectively, from the centre of the sphere. In this process, the magnitude of the total work done on the point charge is $\frac{\rho R^2}{n \epsilon_0}$. The value of $n$ is : ($\epsilon_0$ is the permittivity of vacuum)