Question

(a) In Checkpoint 4, if the dipole rotates from orientation 1 to orientation 2, is the work done on the dipole by the field positive, negative, or zero? (b) If, instead, the dipole rotates from orientation 1 to orientation 4, is the work done by the field more than, less than, or the same as in (a)?

Short answer

a) If the dipole rotates from orientation 1 to 2 in checkpoint 4, the work done on the dipole by the field will be negative.

b) If the dipole rotates from orientation 1 to 4 in checkpoint 4, the work done on the dipole by the field will benegative.

Step-by-step solution

Understanding the concept of work done

The combination of two equal and opposite charges placed at a certain distance is known as an electric dipole. The dipole rotates due to the torque exerted by the electric field on the dipole. This will further result in work done in rotating this dipole.

Formula:

The work done to rotate a dipole in a external electric field,$W=-pE\left(\cos \theta \right)$ (i)

a) Calculation of work done in rotating from 1 to 2

We need to rotate in the clockwise direction to match up with the electric field vector.

Thus, rotating from orientation 1 to 2 clockwise, the work done by the dipole is given using equating (i) as follows:

$\begin{array}{c}{W}_a=W_{orientation2}-W_{orientation1}\\ =-pE\left(\cos \theta \right)-[-pE\left(\cos (\pi -\theta )\right)]\\ =-pE\left(\cos \theta \right)+pE(-\cos \theta )\\ =-2pE\cos \theta \end{array}$

Since, the angle is acute, thus the cosine term will be positive.

Hence, the work done is negative.

b) Calculation of work done in rotating from 1 to 4 

Now, rotating from orientation 1 to 4 clockwise, the work done by the dipole is given using equating (i) as follows:

$\begin{array}{c}{W}_a=W_{orientation4}-W_{orientation1}\\ =-pE\left(\cos (2\pi -\theta )\right)-[-pE\left(\cos (\pi -\theta )\right)]\\ =-pE\left(\cos \theta \right)-pE\left(\cos \theta \right)\\ =-pE\cos \theta \end{array}$$\begin{array}{c}{W}_a=W_{orientation4}-W_{orientation1}\\ =-pE\left(\cos (2\pi -\theta )\right)-[-pE\left(\cos (\pi -\theta )\right)]\\ =-pE\left(\cos \theta \right)-pE\left(\cos \theta \right)\\ =-pE\cos \theta \end{array}$

Hence, the work done in this case will be negative.