$A$ uniform electric field of $400 \, V/m$ is directed $45^\circ$ above the $x$-axis. The potential difference $V_A - V_B$ is: (Coordinates are in cm) (in $, V$)

In a certain region,the electric potential is given by $V = 6x - 8xy^2 - 8y + 6yz - 4z^2 \, V$. The magnitude of the force on a charge of $2 \, C$ at the origin is ........ $N$.

The electric potential $V$ is given as a function of distance $x$ $(m)$ by $V = (2x^2 + 10x - 9) \text{ V}$. The value of the electric field at $x = 1 \text{ m}$ is: (in $\text{ V/m}$)

Two metal plates are separated by $2 \,cm$. The potentials of the plates are $-10 \,V$ and $+30 \,V$. The electric field between the two plates is: (in $\,V/m$)

The electric field in a region of space is given as $E = (5x) \hat{i} \text{ N/C}$. Consider point $A$ on the $Y$-axis at $y = 5 \text{ m}$ and point $B$ on the $X$-axis at $x = 2 \text{ m}$. If the potentials at points $A$ and $B$ are $V_A$ and $V_B$ respectively, then $(V_B - V_A)$ is (in $\text{ V}$)

$A$ uniform electric field having a magnitude $E_0$ and direction along the positive $X$-axis exists. If the potential $V$ is zero at $x = 0$,then its value at $x = +x$ will be: