If $V = -5x + 3y + \sqrt{15}z$,then find the magnitude of the electric field $E(x, y, z)$ in units.

The variation of potential with distance $R$ from a fixed point is as shown in the graph. The electric field at $R = 5\,m$ is......$V/m$.

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?

Two parallel plates are separated by a distance of $5 \, mm$. The potential difference between them is $50 \, V$. $A$ particle of mass $10^{-15} \, kg$ and charge $10^{-11} \, C$ enters the region with a velocity of $10^7 \, m/s$. The acceleration of the particle will be:

In a space having electric field $\vec{E}=A(x \hat{i}+y \hat{j})$,the potential at a point $(10 \ m, 20 \ m)$ is zero. Then the potential at the origin is $\left[A=10 \ Vm^{-2}\right]$. (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