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 

Two parallel plates separated by a distance of $5\,mm$ are kept at a potential difference of $50\,V.$ A particle of mass ${10^{ - 15}}\,kg$ and charge ${10^{ - 11}}\,C$ enters in it with a velocity ${10^7}\,m/s.$ The acceleration of the particle will be

The electrostatic potential inside a charged spherical ball is given by $\phi= a{r^2} + b$ where $r$ is the distance from the centre and $a, b$ are constants. Then the charge density inside the ball 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

In which region magnitude of $x$ -component of electric field is maximum, if potential $(V)$ versus distance $(X)$, graph is as shown?

If on the $x$-axis electric potential decreases uniformly from $60 \,V$ to $20 \,V$ between $x=-2 \,m$ to $x=+2 \,m$, then the magnitude of electric field at the origin