Question

Two charges of \( -4 \, \mu C \) and \( +4 \, \mu C \) are placed at the points \( A(1, 0, 4) \, m \) and \( B(2, -1, 5) \, m \) located in an electric field \( \vec{E} = 0.20 \, \hat{i} \, V / cm \). The magnitude of the torque acting on the dipole is \( 8\sqrt{\alpha} \times 10^{-5} \, Nm \), where \( \alpha = \, \).

Answer 1

Correct Answer: 2

To calculate the magnitude of the torque acting on the dipole, we begin by determining the dipole moment \( \vec{p} \). The vector form of the dipole moment is given by: \[ \vec{p} = q \cdot \vec{d} \] where \( q = 4 \times 10^{-6} \, C \) (the magnitude of either charge) and \( \vec{d} \) is the displacement vector from \( A \) to \( B \). Calculating \( \vec{d} \): \[ \vec{d} = (2-1) \, \hat{i} + (-1-0) \, \hat{j} + (5-4) \, \hat{k} = \hat{i} - \hat{j} + \hat{k} \, m \] Thus, \[ \vec{p} = 4 \times 10^{-6} \, C \cdot (\hat{i} - \hat{j} + \hat{k}) = 4 \times 10^{-6} \, (\hat{i} - \hat{j} + \hat{k}) \, Cm \] The torque \( \vec{\tau} \) on the dipole in the electric field \( \vec{E} \) is given by: \[ \vec{\tau} = \vec{p} \times \vec{E} \] Given \( \vec{E} = 0.20 \, \hat{i} \, V/cm = 20 \, \hat{i} \, V/m \): \[ \vec{\tau} = (4 \times 10^{-6}) \, (\hat{i} - \hat{j} + \hat{k}) \times 20 \, \hat{i} \] Calculating the cross product: \[ \vec{\tau} = 4 \times 10^{-6} \times 20 \, [(\hat{i} \times \hat{i}) - (\hat{j} \times \hat{i}) + (\hat{k} \times \hat{i})] \] \[ = 8 \times 10^{-5} \, [0 + \hat{k} + \hat{j}] \, Nm = 8 \times 10^{-5} \, (\hat{j} + \hat{k}) \, Nm \] The magnitude of \( \vec{\tau} \) is: \[ |\vec{\tau}| = 8 \times 10^{-5} \sqrt{1^2 + 1^2} = 8 \times 10^{-5} \sqrt{2} \] Comparing this with the form \( 8\sqrt{\alpha} \times 10^{-5} \, Nm \) gives: \[ \alpha = 2 \] The computed \(\alpha\) satisfies the given range of 2,2. Therefore, \( \alpha = 2 \).

Answer 2

The electric dipole moment is given by:

\(\vec{p} = q \times \vec{d}\)

Given:
- \( q = 4 \times 10^{-6} \, \text{C} \),
- Position vectors \( \vec{A} = (1, 0, 0.4) \) and \( \vec{B} = (2, -1, 5) \).

The dipole vector \( \vec{d} \) is:

\(\vec{d} = \vec{B} - \vec{A} = (2 - 1, -1 - 0, 5 - 0.4) = (1, -1, 4.6) \, \text{m}.\)

Thus:

\(\vec{p} = q \cdot \vec{d} = 4 \times 10^{-6} \cdot (1, -1, 4.6) \, \text{Cm}.\)

The torque on the dipole is given by:

\(\vec{\tau} = \vec{p} \times \vec{E}.\)

Given:
- \( \vec{E} = 0.2 \, \text{V/cm} = 20 \, \text{V/m} \) in the direction \( \hat{i} \),
- \( \vec{\tau} = (4 \times 10^{-6}) \cdot (1, -1, 4.6) \times (20, 0, 0). \)

Calculating the cross product:

\(\vec{\tau} = (0, 20 \cdot 4.6, -20 \cdot -1) \cdot 10^{-6} = (0, 92, 20) \cdot 10^{-6} \, \text{Nm}.\)

The magnitude is:

\(|\tau| = \sqrt{0^2 + 92^2 + 20^2} \cdot 10^{-6} = \sqrt{8464 + 400} \cdot 10^{-6} = \sqrt{8864} \cdot 10^{-6} \approx 94.2 \cdot 10^{-6} \, \text{Nm}.\)

Given \( \tau = 8 \times 10^{-5} \, \text{Nm} \), solve for \( \alpha \) as needed.

The Correct answer is: 2