$A$ 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:

$A$ spherical shell with an inner radius $a$ and an outer radius $b$ is made of conducting material. $A$ point charge $+Q$ is placed at the centre of the spherical shell and a total charge $-q$ is placed on the shell. Assume that the electrostatic potential is zero at an infinite distance from the spherical shell. The electrostatic potential at a distance $R$ $(a < R < b)$ from the centre of the shell is (where $K = \frac{1}{4\pi\varepsilon_0}$):

The electric potential of the Earth is taken to be zero because the Earth is a good:

Two charges $+q$ and $-q$,each $1 \mu C$,are arranged as shown in the figure. If $x=2 \text{ cm}$ and $y=3 \text{ cm}$,then the potential difference $(V_A - V_B)$ is:

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

$A$ particle has a mass $400$ times that of an electron and a charge double that of an electron. It is accelerated by a potential difference of $5 \ V$. If the particle was initially at rest,what will be its final kinetic energy in $eV$?