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Chapter 20: Problem 60

A dipole with dipole moment \(1.5 \mathrm{nC} \cdot \mathrm{m}\) is oriented at \(30^{\circ}\) to a 4.0-MN/C electric field. Find (a) the magnitude of the torque on the dipole and (b) the work required to rotate the dipole until it's antiparallel to the field.

Short Answer

Expert verified
The torque on the dipole is \(3.0 \mathrm{nNm}\) and the work done to rotate it until it is anti-parallel to the field is \(6.0 \mathrm{nJ}\).

Step by step solution

01

Calculate the torque

Use the formula \(|\tau| = pEsinθ\) to calculate the torque. Here, \(p=1.5 \mathrm{nC} \cdot \mathrm{m}\), \(E=4.0 \mathrm{MN/C}\) and \(θ=30^{\circ}\). So, \(|\tau| = (1.5 \mathrm{nC} \cdot \mathrm{m})(4.0 \mathrm{MN/C})sin30^{\circ} = 3.0 \mathrm{nNm}\).
02

Calculate the work

The work done to rotate the dipole is the change in the potential energy of the dipole from the start to end of the rotation. It is given by \(W = \Delta U = -pEcosθ_1-(-pEcosθ_2)\). Here, \(θ_1=30^{\circ}\) and \(θ_2=180^{\circ}\). This gives \(W = -(1.5 \mathrm{nC} \cdot \mathrm{m})(4.0 \mathrm{MN/C})cos30^{\circ}-[ -(1.5 \mathrm{nC} \cdot \mathrm{m})(4.0 \mathrm{MN/C})cos180^{\circ}] = 6.0 \mathrm{nJ}\).

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