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 $\text{ V}$)

In a region,the potential is represented by $V(x, y, z) = 6x - 8xy - 8y + 6yz$,where $V$ is in volts and $x, y, z$ are in metres. The electric force experienced by a charge of $2 \ C$ situated at point $(1, 1, 1)$ is

$ABC$ 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 $20 \, V$. Given $AB = 3 \, cm$ and $BC = 4 \, cm$. The magnitude of the electric field is (in $S.I.$ units):

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

The electric potential at any point as a function of distance $(x)$ in meters is given by $V = 5x^2 + 10x - 9 \, (V)$. The value of the electric field at $x = 1 \, m$ is ...... $V m^{-1}$.

The figure shows the variation of electric field intensity $E$ versus distance $x$. What is the potential difference between the points at $x = 2 \, m$ and $x = 6 \, m$ from $O$ (in $V$)?