Two charged spheres of radii $10\, cm$ and $15\, cm$ are connected by a thin wire. No current will flow,if they have

$A$ spherical charged conductor has surface charge density $\sigma$. The electric field on its surface is $E$ and the electric potential of the conductor is $V$. Now,the radius of the sphere is halved while keeping the charge constant. The new values of the electric field and potential would be:

As shown in the figure,charges $+q, +q, -q$ and $-q$ are placed on the vertices of a square,each side length is $2l$. The electric potential at the mid-point '$A$' of the charges $+q$ and $+q$ is . . . . . . .

If on the concentric hollow spheres of radii $r$ and $R$ $(R > r)$ the charge $Q$ is distributed such that their surface charge densities are the same,then the potential at their common centre 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?

Two small equal point charges of magnitude $q$ are suspended from a common point on the ceiling by insulating massless strings of equal lengths. They come to equilibrium with each string making an angle $\theta$ from the vertical. If the mass of each charge is $m$,then the electrostatic potential at the centre of the line joining them will be $\left( \frac{1}{4\pi \epsilon_0} = k \right).$