An electric dipole is placed at an angle of $30^{\circ}$ with an electric field of intensity $2 \times…

An electric dipole is placed at an angle of $30^{\circ}$ with an electric field of intensity $2 \times 10^{5} \mathrm{NC}^{-1}$, It experiences a torque of $4 \mathrm{Nm}$. Calculate the charge on the dipole if the dipole length is $2 \mathrm{~cm}$.
  1. $8 \mathrm{mC}$
  2. $4 \mathrm{mC}$
  3. $8 \mu \mathrm{C}$
  4. $2 \mathrm{mC}$

Solution

Torque, $\vec{\tau}=\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{E}}=\mathrm{pE} \sin \theta$
$4=\mathrm{p} \times 2 \times 10^{5} \times \sin 30^{\circ}$
or, $\mathrm{p}=\frac{4}{2 \times 10^{5} \times \sin 30^{\circ}}=4 \times 10^{-5} \mathrm{Cm}$
Dipole moment, $\mathrm{p}=\mathrm{q} \times l$
$q=\frac{p}{l}=\frac{4 \times 10^{-5}}{0.02}=2 \times 10^{-3} \mathrm{C}=2 \mathrm{mC}$ .

Asked in: JEE Mains - Electrostatics - Test 5

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