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 :

$A, B$ and $C$ are three points in a uniform electric field. The electric potential is

Consider a gravity free container as shown. System is initially at rest and electric potential in the regon is $V = (y^3+2)\  J/C$. A ball of charge $q$ and mass $m$ is released from rest from base starts to move up due to electric field and collides with the shaded face as shown.If its speed just after collision is $1.5\  m/s$ and time for which ball is in contact with shaded face is $0.1\ sec$, find external force required to hold the container fixed in its position during collision assuming ball exerts constant force on wall during entire span of collision.......$N$

$A B C$ is a right angled triangle situated in a uniform electric field $\vec{E}$ which is in the plane of the triangle. The points $A$ and $B$ are at the same potential of $15 \,V$ while the point $C$ is at a potential of $20 \,V . A B=3 \,cm$ and $B C=4 \,cm$. The magnitude of electric field is (in $S.I.$ Units)

The electric potential $V$ at any point $(x, y, z),$ all in metres in space is given by $V = 4x^2$ volt. The electric field at the point $(1, 0, 2)$ in volt/meter, is 

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