An electric dipole of dipole moment `10^(-6)` C-m is released from rest in uniform electric field `10^(2)V//m` at angle `theta = 60^(@)`. Maximum rotational kinetic energy of the dipole is say K and maximum torque during the motion is `tau`, then

Options

A. `K = 5.0 xx 10^(-5)J`
B. `K=2.0 xx 10^(-4) J`
C. `tau = 5.0 xx 10^(-4)` N-m
D. `tau = 8.7 xx 10^(-5)` N-m

Answer

To solve the problem, we need to calculate the maximum rotational kinetic energy (K) and the maximum torque (τ) experienced by an electric dipole in a uniform electric field.

### Step 1: Calculate the Potential Energy (U)
The potential energy (U) of an electric dipole in an electric field is given by the formula:
\[ U = -\vec{P} \cdot \vec{E} = -PE \cos(\theta) \]
Where:
- \( P = 10^{-6} \, \text{C-m} \) (dipole moment)
- \( E = 10^2 \, \text{V/m} = 100 \, \text{V/m} \) (electric field)
- \( \theta = 60^\circ \)

Substituting the values:
\[ U = - (10^{-6} \, \text{C-m}) \cdot (100 \, \text{V/m}) \cdot \cos(60^\circ) \]
\[ U = - (10^{-6} \cdot 100) \cdot \frac{1}{2} \]
\[ U = - (10^{-4}) \cdot \frac{1}{2} = -5 \times 10^{-5} \, \text{J} \]

### Step 2: Calculate the Maximum Rotational Kinetic Energy (K)
When the dipole is released from rest, the maximum kinetic energy (K) occurs when the potential energy is at its minimum (which is when the dipole is aligned with the field, \( \theta = 0^\circ \)):
\[ K = U_{\text{initial}} - U_{\text{final}} \]
Where:
- \( U_{\text{initial}} = -5 \times 10^{-5} \, \text{J} \) (calculated above)
- \( U_{\text{final}} = -PE \cos(0^\circ) = -PE \)

Calculating \( U_{\text{final}} \):
\[ U_{\text{final}} = - (10^{-6} \cdot 100) = -10^{-4} \, \text{J} \]

Now substituting into the kinetic energy equation:
\[ K = -5 \times 10^{-5} - (-10^{-4}) \]
\[ K = -5 \times 10^{-5} + 10^{-4} \]
\[ K = 5 \times 10^{-5} \, \text{J} \]

### Step 3: Calculate the Maximum Torque (τ)
The torque (τ) on a dipole in an electric field is given by:
\[ \tau = \vec{P} \times \vec{E} = PE \sin(\theta) \]

Substituting the values:
\[ \tau = (10^{-6} \, \text{C-m}) \cdot (100 \, \text{V/m}) \cdot \sin(60^\circ) \]
\[ \tau = (10^{-6} \cdot 100) \cdot \frac{\sqrt{3}}{2} \]
\[ \tau = 10^{-4} \cdot \frac{\sqrt{3}}{2} \]
\[ \tau \approx 0.87 \times 10^{-4} \, \text{N-m} = 8.7 \times 10^{-5} \, \text{N-m} \]

### Final Answers:
- Maximum Rotational Kinetic Energy (K): \( 5 \times 10^{-5} \, \text{J} \)
- Maximum Torque (τ): \( 8.7 \times 10^{-5} \, \text{N-m} \)