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A teacher models a polar molecule by two charges, +3.0 nC and -3.0 nC, held 2.0 cm apart…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A teacher models a polar molecule by two charges, $+3.0\ \text{nC}$ and $-3.0\ \text{nC}$ , held $2.0\ \text{cm}$ apart. This is much larger than a real molecule, but it keeps the numbers simple. The figure shows two situations.

Two panels. Left panel: a dipole made of a charge of plus 3.0 nanocoulombs on the right and minus 3.0 nanocoulombs on the left, 2.0 cm apart, with the centre O between them. A dashed horizontal line through the charges, the axis, runs to a point P on the right. A dashed vertical line through O, the equatorial line, runs up to a point Q. The distance from O to P and the distance from O to Q are both marked r equal to 0.40 m. Right panel: the same dipole in a uniform field E0 of 5.0 times 10 to the 3 newtons per coulomb drawn as horizontal arrows pointing right. The dipole moment arrow p points up and to the right at 30 degrees to the field.

In the left panel, P lies on the axis of the dipole and Q lies on its equatorial line, and each is $0.40\ \text{m}$ from the centre O. This distance is much larger than the separation of the charges, so you may use the formulae for a short dipole. In the right panel, the same dipole sits in a uniform external field of magnitude $5.0 \times 10^{3}\ \text{N C}^{-1}$ , with its dipole moment at $30^\circ$ to the field. Take $\dfrac{1}{4\pi\varepsilon_0} = 9.0 \times 10^{9}\ \text{N m}^2\text{C}^{-2}$ .

What is the magnitude of the dipole moment?
What is the magnitude of the field at P?
What is the magnitude of the field at Q?
What is the magnitude of the torque on the dipole in the right panel?
How much work must an external agent do to turn the dipole slowly from $\theta = 0^\circ$ to $\theta = 120^\circ$ ?

Decide which of P and Q has the stronger field before you calculate, and watch the sign of the work.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Given. $q = 3.0\ \text{nC}$ , separation $2a = 2.0\ \text{cm}$ , $r = 0.40\ \text{m}$ , $E = 5.0 \times 10^{3}\ \text{N C}^{-1}$ .

(a) Dipole moment. $p = q \times 2a = 3.0 \times 10^{-9}\ \text{C} \times 0.020\ \text{m} = 6.0 \times 10^{-11}\ \text{C m}$ .

(b) On the axis. $E = \dfrac{2kp}{r^3}$ , with $k = 9.0 \times 10^{9}\ \text{N m}^2\text{C}^{-2}$ . Here $2kp = 1.08\ \text{N m}^3\text{C}^{-1}$ and $r^3 = 0.064\ \text{m}^3$ , so $E = 16.9\ \text{N C}^{-1}$ .

(c) On the equatorial line. $E = \dfrac{kp}{r^3} = 8.4\ \text{N C}^{-1}$ , half of the field on the axis. It points opposite to $\vec p$ .

(d) Torque. $\tau = pE\sin\theta = 6.0 \times 10^{-11} \times 5.0\times10^{3} \times \sin 30^\circ = 1.5 \times 10^{-7}\ \text{N m}$ .

(e) Work. $U = -pE\cos\theta$ , so $W = U(120^\circ) - U(0^\circ) = pE(1 - \cos 120^\circ) = pE \times 1.5 = 4.5 \times 10^{-7}\ \text{J}$ . The work is positive: the agent works against the field.

Answer: (a) $6.0 \times 10^{-11}\ \text{C m}$ . (b) $16.9\ \text{N C}^{-1}$ . (c) $8.4\ \text{N C}^{-1}$ . (d) $1.5 \times 10^{-7}\ \text{N m}$ . (e) $4.5 \times 10^{-7}\ \text{J}$ .

Check. The exact fields of the two charges at $0.40\ \text{m}$ are $16.9\ \text{N C}^{-1}$ on the axis and $8.4\ \text{N C}^{-1}$ on the equatorial line, the same to one decimal place. So the short-dipole formulae are good here.

Watch out. The ratio of the two fields is $2$ , whatever the distance. If you use $W = pE\cos 120^\circ$ you get $-1.5 \times 10^{-7}\ \text{J}$ , a negative number: that is the work done by the field, not by the agent.

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