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
Electric Field Intensity: The electric field 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.,$\vec{\text{E}}=\lim\limits_{\text{q}_0\rightarrow0}\frac{\vec{\text{F}}}{\text{q}_0}$
The test charge $q0$ 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 mid $-$ point of dipole: The electric field strength at mid $-$ point $C$ due to charge $+q$ is $-q$ along the same direction.
$E = E_1 + E_2=\frac{1}{4\pi\epsilon_0}\frac{\text{q}}{\text{a}^2}+\frac{1}{4\pi\epsilon_0}\frac{\text{q}}{\text{a}^2}=\frac{1}{4\pi\epsilon_0}\frac{2\text{q}}{\text{a}^2}$
Its direction is from $+q$ to $-q$.
i.e.,$\vec{\text{E}}=\lim\limits_{\text{q}_0\rightarrow0}\frac{\vec{\text{F}}}{\text{q}_0}$
The test charge $q0$ 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 mid $-$ point of dipole: The electric field strength at mid $-$ point $C$ due to charge $+q$ is $-q$ along the same direction.
$E = E_1 + E_2=\frac{1}{4\pi\epsilon_0}\frac{\text{q}}{\text{a}^2}+\frac{1}{4\pi\epsilon_0}\frac{\text{q}}{\text{a}^2}=\frac{1}{4\pi\epsilon_0}\frac{2\text{q}}{\text{a}^2}$
Its direction is from $+q$ to $-q$.
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