Electric Field MCQ - Practice Questions with Answers

Electric field

Electric field and electric lines of force:

The space around a charge in which another charged particle experiences a force is said to have an electric field in it. 

Electric Field Intensity: 

The electric field intensity at any point is defined as the force experienced by a unit positive charge placed at that point.

Electric field intensity,

$E=\frac{F}{q_0}$

where F is the force experienced by q 0 . The SI unit of E is,

$\frac{\text{ Newton }}{\text{ Columb }}=\frac{\text{ Volt }}{\text{ meter }}=\frac{\text{ Joule }}{\text{ Coulomb }\times \text{ Meter }}$

The dimensional formula is $\left[ML{T}^{-3}{A}^{-1}\right]$

The electric field is a vector quantity and due to positive charge is away from the charge and for the negative charge, it is towards the charge.

Electric field due to a point charge: 

Consider a point charge placed at the origin $O$. Let a test charge $q0$ is placed at $P$ which is at a distance $r$ from $O$. Force $F$ on test charge $q0$ is

$F=\frac{1}{4\pi {ϵ}_0}\frac{q{q}_0}{r^2}\hat{r}$

The electric field at point $P$ due to $q$ is

$E=\underset{q_0\to 0}{lim}\frac{F}{q_0}=\underset{q_0\to 0}{lim}\frac{1}{q_0}\frac{1}{4\pi {ϵ}_0}\frac{q{q}_0}{r^2}\hat{r}=\frac{1}{4\pi {ϵ}_0}\frac{q}{r^2}\hat{r}$

(As ${\underset{―}{q}}_{\underset{―}{o}}$ tends to zero the electric field produced by $q$ is not affected by ${\underset{―}{q}}_{\underset{―}{\underset{―}{o}}}$.)
The magnitude of the electric field

$E=\frac{1}{4\pi {ϵ}_0}\frac{q}{r^2}$

Electric field due to a system of charge:

Electric field obeys the superposition principle. That is the electric field due to a system of charge at a point is equal to the vector sum of all the electric fields.