Question

In a uniform electric field $10\text{ N C}^{-1}$, an electric dipole of length $4\text{ cm}$ is placed with its axis making an angle $60^\circ$ with the electric field. If the dipole experiences a torque of $8\sqrt{3}\text{ N m}$, find the potential energy of the dipole.

• A. $8\text{ J}$
• B. $-8\text{ J}$
• C. $-16\text{ J}$
• D. $16\text{ J}$

Hint:

You can directly link torque and potential energy for any system when given the same angle by using the simple identity: $U = -\tau \cot\theta$. Here, $U = -(8\sqrt{3}) \cot(60^\circ) = -8\sqrt{3} \times \frac{1}{\sqrt{3}} = -8\text{ J}$. This completely bypasses finding $p$ or $E$ explicitly!

Answer 1

The Correct Option is B

Understanding the Concept: An electric dipole placed in a uniform electric field experiences a turning torque ($\tau$) given by the vector cross product $\vec{\tau} = \vec{p} \times \vec{E}$, whose magnitude is: \[ \tau = pE\sin\theta \] The potential energy ($U$) stored in the system due to this orientation is given by the dot product $U = -\vec{p} \cdot \vec{E}$, which expands to: \[ U = -pE\cos\theta \]
Step 1: Calculate the value of $pE$ using the torque formula.
We are given:
Electric field, $E = 10\text{ N C}^{-1}$
Angle, $\theta = 60^\circ$
Torque, $\tau = 8\sqrt{3}\text{ N m}$
Substitute these values into the torque expression: \[ 8\sqrt{3} = pE \sin(60^\circ) \] We know that $\sin(60^\circ) = \frac{\sqrt{3}}{2}$. Placing this in the equation: \[ 8\sqrt{3} = pE \left(\frac{\sqrt{3}}{2}\right) \implies pE = 8 \times 2 = 16\text{ N m} \]
Step 2: Determine the potential energy ($U$).
Now, substitute the value of $pE = 16$ and $\theta = 60^\circ$ into the potential energy formula: \[ U = -pE\cos(60^\circ) \] Since $\cos(60^\circ) = \frac{1}{2}$: \[ U = -16 \times \frac{1}{2} = -8\text{ J} \]