Electric potential is given by $V = 6x - 8xy^2 - 8y + 6yz - 4z^2$. Then,the electric force acting on a $2 \, C$ point charge placed at the origin will be......$N$.

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

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......$V/m$.

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 $V$)?

$A$ charge of $5\,C$ experiences a force of $5000\,N$ when it is kept in a uniform electric field. What is the potential difference $V$ (in volts) between two points separated by a distance of $1\,cm$?

Assume that an electric field $\vec{E} = 30x^2 \hat{i}$ exists in space. Then the potential difference $V_A - V_O$,where $V_O$ is the potential at the origin and $V_A$ is the potential at $x = 2 \ m$,is....$V$