The electric potential at a point $(x, y, z)$ is given by $V=-x^2y-xz^3 +4 $. The electric field at that point is
$\vec E = 2xy\hat i + \left( {{x^2} + {y^2}} \right)\hat j + \left( {3xz - {y^2}} \right)\hat k$
$\;\vec E = {z^3}\hat i + xyz\hat j + {z^2}\hat k$
$\;\vec E = \left( {2xy - {z^3}} \right)\hat i + x{y^2}\hat j + 3{z^2}x\hat k$
$\;\vec E = \left( {2xy + {z^3}} \right)\hat i + {x^2}\hat j + 3x{z^2}\hat k$
At a certain distance from a point charge the electric field is $500\,V/m$ and the potential is $3000\,V$. What is this distance......$m$
A sphere carrying charge of $Q$ having weight $w$ falls under gravity between a pair of vertical plates at a distance of $d$ from each other. When a potential difference $V$ is applied between the plates the acceleration of sphere changes as shown in the figure, to along line $BC$. The value of $Q$ is :-
Variation of electrostatic potential along $x$-direction is shown in the graph. The correct statement about electric field is
Two plates are $2\,cm$ apart, a potential difference of $10\;volt$ is applied between them, the electric field between the plates is.........$N/C$
$A, B$ and $C$ are three points in a uniform electric field. The electric potential is