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

The electrostatic potential inside a charged spherical ball is given by : $V = b -ar^2$, where  $r$ is the distance from the centre ; $a$ and $b$ are constants. Then, the charge density inside the ball is :

The potential (in volts ) of a charge distribution is given by

$V(z)\, = \,30 - 5{z^2}for\,\left| z \right| \le 1\,m$

$V(z)\, = \,35 - 10\,\left| z \right|for\,\left| z \right| \ge 1\,m$

$V(z)$  does not depend on $x$  and  $y.$  If this potential is generated by a constant charge per unit volume $\rho _0$ (in units of $\varepsilon _0$ ) which is spread over a certain region, then choose the correct statement

The maximum electric field that can be held in air without producing ionisation of air is $10^7\,V/m$. The maximum potential therefore, to which a conducting sphere of radius $0.10\,m$ can be charged in air is

The electric potential $V(x)$ in a region around the origin is given by $V(x) = 4x^2\,volts$ . The electric charge enclosed in a cube of $1\,m$ side with its centre at the origin is (in coulomb)

The electric potential at a point in free space due to charge $Q$ coulomb is  $V=Q$$ \times {10^{11}}\,V$ . The electric field at that point is