Question

An electric dipole of length $2cm$, when placed with its axis making an angle of $\theta =60$ with a uniform electric field, experiences a torque of $8\sqrt{3}Nm$. Calculate the potential energy of the dipole, if it has a charge of $\pm 4nC.$

Answer 1

Step1: Given data

1. The length of the dipole is $L=2cm.$ .
2. The torque exerted on the dipole is $\tau =$$8\sqrt{3}Nm$.
3. The charges of the dipole are $\pm 4nC.$.
4. The dipole is placed at an angle with the electric field is $\theta =60$.

Step2: Dipole in a uniform electric field

1. When two opposite charges of the same value are placed at a distance, then the system is called a dipole.
2. So, any dipole consists of two charges, $+q$ and $-q$ .
3. Dipole moment of a dipole is defined by the form, $P=qL$, where q is the charge and L is the length of the dipole.
4. We know in an electric field, the torque on a dipole is $\tau =PE\sin \theta$, where P is the dipole moment and E is the uniform electric field.
5. The work done to rotate a dipole at an angle $\theta =60$ in a uniform electric field is, $W={\int }_{{\theta }_1}^{{\theta }_2}PE\sin \theta .d\theta$ .This work done is stored as potential energy in the dipole, when it is rotated at an angle $d\theta$.

Step3: Diagram

Step4: Calculation of dipole moment

We know dipole is defined by the form, $P=qL$

So, $$\begin{gathered} P=\left(2cm\right)\times \left(4nC\right)=\left(2\times {10}^{-2}\right)\times \left(4\times {10}^{-9}\right) \\ orP=8\times {10}^{-11}Cm................\left(1\right) \end{gathered}$$

Step5: Calculation of electric field

As we know torque, $\tau =PE\sin \theta$

$$\begin{gathered} or,E=\frac{\tau }{P\sin \theta }=\frac{8\sqrt{3}}{\left(8\times {10}^{-11}\right)\times {\displaystyle \frac{\sqrt{3}}{2}}}=2\times {10}^{11} \\ orE=2\times {10}^{11}Nm.......................\left(2\right) \end{gathered}$$

Step6: The potential energy of the dipole in an electric field

As we know that the potential energy of the dipole when it is placed at an angle $d\theta$ with the electric field is $U={\int }_{{\theta }_1}^{{\theta }_2}PE\sin \theta .d\theta$

Now, the potential energy of the dipole at the position $\theta =60$ is,

$$\begin{gathered} U={\int }_{{\theta }_1}^{{\theta }_2}PE\sin \theta .d\theta =PE{\int }_{{\theta }_1}^{{\theta }_2}\sin \theta .d\theta \\ orU=\left(8\times {10}^{-11}\right)\times \left(2\times {10}^{11}\right)\times {\left[-\cos \theta \right]}_0^{60} \\ orU=-16\times \left(\frac{1}{2}-1\right) \\ orU=8Joule. \end{gathered}$$ (using equations 1 and 2)

Therefore, the potential energy of the dipole is $8joule.$