When a charge of $0.01 \ C$ is moved from point $A$ to point $B$ against an electric field,the work done is $15 \ J$. The potential difference $(V_B - V_A)$ is ....... $V$.

$A$ charge of total amount $Q$ is distributed over two concentric hollow spheres of radii $r$ and $R$ $(R > r)$ such that the surface charge densities on the two spheres are equal. The electric potential at the common centre is

$A$ small conducting sphere of radius $r$ is lying concentrically inside a bigger hollow conducting sphere of radius $R$. The bigger and smaller spheres are charged with $Q$ and $q$ respectively $(Q > q)$ and are insulated from each other. The potential difference between the spheres will be

Along the $x$-axis,three charges $\frac{q}{2}, -q$ and $\frac{q}{2}$ are placed at $x=0, x=a$ and $x=2a$ respectively. The resultant electric potential at a point $P$ located at a distance $r$ from the charge $-q$ $(r > a)$ is ($\varepsilon_0$ is the permittivity of free space):

Let $A, B$ and $C$ be the three points in a uniform electric field $(\overrightarrow{E})$ as shown. The electric potential is

$A$ regular hexagon of side $10 \text{ cm}$ has a charge of $1 \mu\text{C}$ at each of its vertices. The potential at the centre of the hexagon is $\left[\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ SI unit}\right]$.