Previous Year NEET Question- Electrostatics L-4

Question: 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 equal to $4 \mathrm{~N} \mathrm{~m}$. Calculate the magnitude of charge on the dipole, if the dipole length is $2 \mathrm{~cm}$. (NEET-2023)#

A) 6 mC

B) 4 mC

C) 2 mC

D) 8 mC

Answer: 2 mC#

Explanation#

The torque $\tau$ experienced by an electric dipole in an electric field is given by the formula: $$ \tau=p E \sin \theta $$ where $\mathrm{p}$ is the electric dipole moment, $\mathrm{E}$ is the electric field intensity, and $\theta$ is the angle between the dipole and the electric field. The electric dipole moment p can be expressed as: $$ p=q d $$ where $\mathrm{q}$ is the charge on the dipole, and $\mathrm{d}$ is the dipole length.

We are given the following values:

Torque $\tau=4 \mathrm{~N} \cdot \mathrm{m}$ Electric field intensity $\mathrm{E}=2 \times 10^5 \mathrm{NC}^{-1}$ Angle $\theta=30^{\circ}$ Dipole length $d=2 \mathrm{~cm}=0.02 \mathrm{~m}$

We need to find the charge $q$ on the dipole. Let’s first solve for the electric dipole moment $\mathrm{p}$ : $\tau=p E \sin \theta$ $$ \Rightarrow p=\frac{\tau}{E \sin \theta} $$

Substituting the given values: $$ p=\frac{4}{\left(2 \times 10^5\right) \sin 30^{\circ}}=\frac{4}{\left(2 \times 10^5\right)(0.5)}=\frac{4}{10^5}=4 \times 10^{-5} \mathrm{C} \mathrm{m} $$

Now, let’s solve for the charge q using the formula:

$\Rightarrow p=q d$

$q=\frac{p}{d}$

Substituting the values for $p$ and $d$ :

$q=\frac{4 \times 10^{-5}}{0.02}=2 \times 10^{-3} \mathrm{C}=2 \mathrm{mC}$

So, the magnitude of the charge on the dipole is $2 \mathrm{mC}$.