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)

Figure shows three points $A$, $B$ and $C$ in a region of uniform electric field $\overrightarrow E $. The line $AB$ is perpendicular and $BC$ is parallel to the field lines. Then which of the following holds good. Where ${V_A} > {V_B}$ and ${V_C}$ represent the electric potential at points $A$, $B$ and $C$ respectively

$A B C$ 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 at a point $(x,\;y)$ in the $x - y$ plane is given by $V = - kxy$. The field intensity at a distance $r$ from the origin varies as

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 :

In a certain region of space, variation of potential with distance from origin as we move along $x$-axis is given by $V=8 x^2+2$, where $x$ is the $x$-coordinate of a point in space. The magnitude of electric field at a point $(-4,0)$ is .......... $V / m$