When a $5 \text{ C}$ charge is placed in a uniform electric field, a force of $5000 \text{ N}$ acts on it. Find the potential difference between two points in that field, separated by a distance of $1 \text{ cm}$. (in $\text{V}$)

Electric potential at any point is given by $V = -5x + 3y + \sqrt{15}z$. The magnitude of the electric field is:

The electrostatic potential inside a charged spherical ball is given by $V = b - ar^2$,where $r$ is the distance from the centre; $a$ and $b$ are constants. Then,the charge density inside the ball is:

The electric potential $(V)$ as a function of distance $(x)$ [in meters] is given by $V = (5x^2 + 10x - 9) \, V$. The value of the electric field at $x = 1 \, m$ would be ...... $V/m$.

The potential (in volts) of a charge distribution is given by $V(z) = 30 - 5z^2$ for $|z| \le 1 \ m$ and $V(z) = 35 - 10|z|$ for $|z| \ge 1 \ m$. $V(z)$ does not depend on $x$ and $y$. If this potential is generated by a constant charge per unit volume $\rho_0$ (in units of $\varepsilon_0$) which is spread over a certain region,then choose the correct statement.

If the electric potential at any point $(x, y, z) \, m$ in space is given by $V = 3x^2$ volt,the electric field at the point $(1, 0, 3) \, m$ will be ............