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.
$E = E_1 + E_2=\frac{1}{4\pi\in_0}\frac{\text{q}}{\text{a}^2}+\frac{1}{4\pi\in_0}\frac{\text{q}}{\text{a}^2}=\frac{1}{4\pi\in_0}\frac{2\text{q}}{\text{a}^2}$
Its direction is from +q to -q.
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.
$E = E_1 + E_2=\frac{1}{4\pi\in_0}\frac{\text{q}}{\text{a}^2}+\frac{1}{4\pi\in_0}\frac{\text{q}}{\text{a}^2}=\frac{1}{4\pi\in_0}\frac{2\text{q}}{\text{a}^2}$
Its direction is from +q to -q.
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