The electric potential energy of a $2 \mu C$ charge is $3 \times 10^{-5} \text{ J}$ at a point in a uniform electric field. The electric potential at the point is . . . . . . $V$.

The potential difference $(V_A-V_B)$ in the arrangement shown in the figure is $(q=1 \mu C, x=2 \text{ cm}, y=3 \text{ cm})$.

$A$ 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)

$A$ particle of mass $m$ and charge $q$ is accelerated through a potential difference of $V$ volt. Its energy is

Two point charges $-q$ and $+q$ are located at points $(0,0,-a)$ and $(0,0, a)$ respectively. The electric potential at a point $(0,0, z)$,where $z>a$ is

Explain electrostatic potential energy difference and give the noteworthy comments on it.