Question

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

Answer 1

Hint: When a charged particle is kept near to the other charged particle, both the charges experience a force. It may be attractive if the charges are of opposite nature and it may be repulsive if the charges are alike. The charge which is brought near to the other charge is called a test charged particle and the other charge is called a source charged particle. The electric field intensity tells us about the magnitude and direction of the force experienced by the charged particle.
When the two equal charges but of opposite nature are kept at a very short distance the system of two charged particles is called an electric dipole.

Complete step by step answer:
Step 1:
Definition- the electric field intensity is defined as the ratio of the force $\overrightarrow{F}$ experienced by the charge and the amount of the charge $q$ .
$\therefore \overrightarrow{E}={\displaystyle \frac{\overrightarrow{F}}{q}}$
Force is a vector quantity.
The SI unit of the electric field intensity is volt/meter or V/m.
The coulomb’s law says that the force between two charges is given by $F=k{\displaystyle \frac{qQ}{r^2}}$ where $k={\displaystyle \frac{1}{4\pi {\epsilon }_0}}$ is a constant and $r$ is the distance from the source charge. Therefore, the above equation can be written as
$\therefore \overrightarrow{E}=k{\displaystyle \frac{qQ}{r^2}}{\displaystyle \frac{1}{q}}$
$\Rightarrow \overrightarrow{E}=k{\displaystyle \frac{Q}{r^2}}$
$Q$ is the source charge.

Step 2:
The length of the electric dipole is $2a$ therefore the distance of the midpoint from each charged particle is $a$ . In the middle of the dipole, the electric fields contributed by both charges are in the same direction; they are both pointing toward the negative charge. Thus, the magnitude of the electric field is the sum of the contribution of the two charges. Let the amount of charge be $q$ .
$\therefore E=k{\displaystyle \frac{q}{a^2}}+k{\displaystyle \frac{q}{a^2}}$
$\Rightarrow E=k{\displaystyle \frac{2q}{a^2}}$, where $k={\displaystyle \frac{1}{4\pi {\epsilon }_0}}$
This is the electric field intensity due to the electric dipole of length $2a$ .

Note: The derivation above shows that the test charge $q$ was canceled from both the numerator and denominator of the equation. The new formula for electric field intensity expresses the field strength in terms of the two variables that affect it. The electric field strength is dependent upon the quantity of charge on the source charge $Q$ and the distance of separation $r$ from the source charge.