If the potential energy of an electron and a proton at a distance $r$ is given by the formula $U = - \left( \frac{ke^2}{3r^3} \right)$,then which law of force applies?

The electric field in a region is given by $\vec{E} = (Ax + B)\hat{i}$ where $E$ is in $N\,C^{-1}$ and $x$ is in meters. The values of constants are $A = 20\, SI\, \text{unit}$ and $B = 10\, SI\, \text{unit}$. If the potential at $x = 1$ is $V_1$ and that at $x = -5$ is $V_2$, then $V_1 - V_2$ is.....$V$.

The variation of potential $V$ with distance $x$ from a fixed point is as shown in the figure. The electric field at $x = 13\,m$ is......$V/m$.

The potential function of an electrostatic field is given by $V = 2x^2$. Determine the electric field strength at the point $(2 \ m, 0, 3 \ m)$.

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: