Question

Define electric field intensity. Write its $SI$unit. Write the magnitude and direction of electric field intensity due to a dipole of length $2a$at the midnight of the line joining the two charges.

Answer 1

Hint: The electric field is a region around a charge in which it exerts an electrostatic force on other charges. While the strength of the electric field at any point in space is called electric field intensity. It is a vector quantity.

Complete step by step answer:

Electric field intensity: the electric intensity at any point in an electric field is defined as the electric force per unit positive test charge placed at that point i.e.
$\vec E = \mathop {\lim }\limits_{{q^{ \to 0}}} \dfrac{{\vec F}}{{{q_0}}}$
The test charges ${q_0}$has to be vanishingly small so that it does not affect the electric field of the main charge.
The $SI$ unit of electric field intensity is newton/coulomb.
Electric field strength at midpoint of dipole: the electric field strength at midpoint $C$due to charge $ + q$is $ - q$along the same direction.
$E = {E_1} + {E_2} = \dfrac{1}{{4\pi { \in _0}}}\dfrac{q}{{{a^2}}} + \dfrac{1}{{4\pi { \in _0}}}\dfrac{q}{{{a^2}}} $
$= \dfrac{1}{{4\pi { \in _0}}}\dfrac{{2q}}{{{a^2}}}$
Its direction is from $ + q$to $ - q$.

Note:
A measure of the force exerted by one charged body on another. The electric field intensity (volts/meter) at any location is the force (Newtons) that would be experienced by unit test charge placed at the location. Electric field is the negative gradient of the scalar potential. The negative sign came as a result because the potential difference is the work done per unit charge against the electrostatic force to move a charge from $a$to $b$. The electric field intensity at any point due to a system or group of charges is equal to the vector sum of electric field intensities due to individual charges at the same point.