The electric potential at any point $(x, y, z)$ in metres is given by $V = 3x^2$. The electric field at a point $(2, 0, 1)$ is (in $Vm^{-1}$)

$A$ uniform electric field of $20\, N/C$ exists along the $x$-axis in a space. The potential difference $(V_B - V_A)$ for the points $A(4\,m, 2\,m)$ and $B(6\,m, 5\,m)$ is.....$V$.

Assume that an electric field $E = 30x^2 \hat{i}$ exists in space. If $V_0$ is the potential at the origin and $V_A$ is the potential at $x = 2 \ m$,then the potential difference $(V_A - V_0)$ is: (in $V$)

As shown in the figure, if the values of the electric potential at three points $A, B$ and $C$ in a uniform electric field $(\vec{E})$ are $V_A, V_B$, and $V_C$ respectively, then

The electric potential at a place is varying as $V = \frac{1}{2}(y^2 - 4x) \text{ V}$. Then the electric field at $x = 1 \text{ m}$ and $y = 1 \text{ m}$ is

$A$ small particle carrying a negative charge of $1.6 \times 10^{-19} \text{ C}$ is suspended in equilibrium between two horizontal metal plates $8 \text{ cm}$ apart having a potential difference of $980 \text{ V}$ across them. Find the mass of the particle. $[g = 9.8 \text{ m/s}^2]$