What is the electric potential at point $A$ due to the given charges?

The electrostatic potential on the surface of a uniformly charged spherical shell of radius $R = 10 \ cm$ is $120 \ V$. The potential at the centre of the shell,at a distance $r = 5 \ cm$ from the centre,and at a distance $r = 15 \ cm$ from the centre of the shell respectively,are:

Four charges are arranged at the corners of a square $ABCD$ of side $d$,as shown in Figure.
$(a)$ Find the work required to put together this arrangement.
$(b)$ $A$ charge $q_{0}$ is brought to the centre $E$ of the square,the four charges being held fixed at its corners. How much extra work is needed to do this?

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

Charges of $1 \mu C$ are placed at each of the four corners of a square of side $2 \sqrt{2} \text{ m}$. The potential at the point of intersection of the diagonals is . . . . . . .

$A$ point charge $-q$ is carried from a point $A$ to another point $B$ on the axis of a charged ring of radius $r$ carrying a charge $+q$. If the point $A$ is at a distance $\frac{4}{3} r$ from the centre of the ring and the point $B$ is $\frac{3}{4} r$ from the centre but on the opposite side, what is the net work that needs to be done for this?