Electric Field - Practice Questions & MCQ

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 Sl 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 $\left[M L T^{-3} A^{-1}\right]$

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 \epsilon_0} \frac{q q_0}{r^2} \hat{r}
$

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

$
E=\lim _{q_0 \rightarrow 0} \frac{F}{q_0}=\lim _{q_0 \rightarrow 0} \frac{1}{q_0} \frac{1}{4 \pi \epsilon_0} \frac{q q_0}{r^2} \hat{r}=\frac{1}{4 \pi \epsilon_0} \frac{q}{r^2} \hat{r}
$

(As $\mathrm{q}_0$ tends to zero the electric field produced by q is not affected by $\mathrm{q}_0$.)
The magnitude of the electric field

$
E=\frac{1}{4 \pi \epsilon_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.