The electric potential $V$ is given as a function of distance $x$ (metre) by $V = (5{x^2} + 10x - 9)\,volt$. Value of electric field at $x = 1$ is......$V/m$

For a charged spherical ball, electrostatic potential inside the ball varies with $r$ as $V =2 ar ^2+ b$. Here, $a$ and $b$ are constant and $r$ is the distance from the center. The volume charge density inside the ball is $-\lambda a \varepsilon$. The value of $\lambda$ is $...........$. $\varepsilon=$ permittivity of medium.

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 $2$ coulomb situated at point $( 1, 1, 1)$ is

At a certain distance from a point charge the electric field is $500\,V/m$ and the potential is $3000\,V$. What is this distance......$m$

The potential due to an electrostatic charge distribution is $V(r)=\frac{q e^{-\alpha e r}}{4 \pi \varepsilon_{0} r}$, where $\alpha$ is positive. The net charge within a sphere centred at the origin and of radius $1/ \alpha$ is

The electrostatic potential inside a charged spherical ball is given by $\phi= a{r^2} + b$ where $r$ is the distance from the centre and $a, b$ are constants. Then the charge density inside the ball is: