Solveeit Logo
Login

Question

Physics Question on Electrostatics

Two charges of −4 μC-4 \, \mu C and +4 μC+4 \, \mu C are placed at the points A(1,0,4) mA(1, 0, 4) \, m and B(2,−1,5) mB(2, -1, 5) \, m located in an electric field E⃗=0.20 i^ V/cm\vec{E} = 0.20 \, \hat{i} \, V / cm. The magnitude of the torque acting on the dipole is 8α×10−5 Nm8\sqrt{\alpha} \times 10^{-5} \, Nm, where α= \alpha = \,.

Answer

The electric dipole moment is given by:

p⃗=q×d⃗\vec{p} = q \times \vec{d}

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

The dipole vector d⃗\vec{d} is:

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

Thus:

p⃗=q⋅d⃗=4×10−6⋅(1,−1,4.6) Cm.\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:

τ⃗=p⃗×E⃗.\vec{\tau} = \vec{p} \times \vec{E}.

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

Calculating the cross product:

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

The magnitude is:

∣τ∣=02+922+202⋅10−6=8464+400⋅10−6=8864⋅10−6≈94.2⋅10−6 Nm.|\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 τ=8×10−5 Nm\tau = 8 \times 10^{-5} \, \text{Nm}, solve for α\alpha as needed.

The Correct answer is: 2