The election field in a region is given by $\vec E = (Ax + B)\hat i$ where $E$ is in $N\,C^{-1}$ and $x$ in meters. The values of constants are $A = 20\, SI\, unit$ and  $B = 10\, SI\, unit$. If the potential at $x =1$ is $V_1$ and that at $x = -5$ is $V_2$ then $V_1 -V_2$ is.....$V$

Three concentric metallic spherical shell $A, B$ and $C$ or radii $a, b$ and $c$ $(a < b < c)$ have surface charge densities $- \sigma , + \sigma ,$ and $- \sigma $ respectively. The potential of shell $A$ is :

Two thin concentric hollow conducting spheres of radii $R_1$ and $R_2$ bear charges $Q_1$ and $Q_2$ respectively. If $R_1 < R_2$, then the potential of a point at a distance $r$ from the centre $(R_1 < r < R_2)$ is

If the potential at the centre of a uniformly charged hollow sphere of radius $R$ is $V$ then electric field at a distance $r$ from the centre of the sphere is $(r > R)$

The electric potential at the surface of an atomic nucleus $(Z = 50)$ of radius $9.0×{10^{ - 13}}\, cm$ is

Charges are placed on the vertices of a square as shown Let $\vec E$ be the electric field and $V$ the potential at the centre. If the charges on $A$ and $B$ are interchanged with those on $D$ and $C$ respectively, then