Q.Define intensity of electric field at a point. Derive an expression for the intensity due to a point charge.

Electric field intensity is force per unit test charge; applying Coulomb's law to a small positive test charge near a point charge $Q$ gives $E = kQ/r^2$.

Electric field intensity

The electric field intensity (or electric field) at a point in space is defined as the electrostatic force experienced per unit positive test charge placed at that point, in the limit that the test charge is vanishingly small (so it doesn't disturb the source charge distribution):

$\vec{E} = \lim_{q_0 \to 0} \dfrac{\vec{F}}{q_0}$

It is a vector quantity, with SI unit N/C (equivalently V/m), and its direction is the direction of the force on a positive test charge.

Electric field due to a point charge

Consider a point charge $Q$ placed at the origin. Let a small positive test charge $q_0$ be placed at a point P at a distance $r$ from $Q$. By Coulomb's law, the force experienced by $q_0$ is:

$F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q q_0}{r^2}$

The electric field at P is then:

$E = \dfrac{F}{q_0} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r^2}$

In vector form, with $\hat{r}$ the unit vector from $Q$ towards P: …