Consider a sphere of radius $R$ having charge $q$ uniformly distributed inside it. At what minimum distance from its surface the electric potential is half of the electric potential at its centre?

If a charged spherical conductor of radius $10\,cm$ has potential $V$ at a point distant $5\,cm$ from its centre, then the potential at a point distant $15\,cm$ from the centre will be

A charge $+q$ is fixed at each of the points $x = x_0,\,x = 3x_0,\,x = 5x_0$, .... upto $\infty $ on $X-$  axis and charge $-q$ is fixed on each of the points $x = 2x_0,\,x = 4x_0,\,x = 6x_0$, .... upto $\infty $ . Here $x_0$ is a positive constant. Take the potential at a point due to a charge $Q$ at a distance $r$ from it to be $\frac{Q}{{4\pi {\varepsilon _0}r}}$. Then the potential at the origin due to above system of charges will be

The charge given to a hollow sphere of radius $10\, cm$ is $3.2×10^{-19}\, coulomb$. At a distance of $4\, cm$ from its centre, the electric potential will be

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

The electric field $\vec E$ between two points is constant in both magnitude and direction. Consider a path of length d at an angle $\theta = 60^o$ with respect to field lines shown in figure. The potential difference between points $1$ and $2$ is