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?

As shown in the figure,charges $+q$ and $-q$ are placed at the vertices $B$ and $C$ of an isosceles triangle. The potential at the vertex $A$ is

Define $1 \ eV$.

Two charges $q_1$ and $q_2$ are separated by a distance of $30\ cm$. $A$ third charge $q_3$,initially at $C$ as shown in the figure,is moved along the circular path of radius $40\ cm$ from $C$ to $D$. If the difference in potential energy due to the movement of $q_3$ from $C$ to $D$ is given by $\frac{q_3 K}{4 \pi \epsilon_0}$,find the value of $K$.

As shown in the figure,on bringing a charge $Q$ from point $A$ to $B$ and from $B$ to $C$,the work done is $2 \, J$ and $-3 \, J$ respectively. The work done to bring the charge from $C$ to $A$ is (in $, J$)