Question

An electric dipole with dipole moment $$\overrightarrow{p}=\left(3\hat{i}+4\hat{j}\right)\times {10}^{-30}\text{C}\cdot \text{m}$$ is placed in an electric field $$\overrightarrow{E}=4000\hat{i}\text{N}\cdot \text{c}$$. An external agent turns the dipole and is dipole moment becomes $$\left(-4\hat{i}+3\hat{j}\right)\times {10}^{-30}\text{C}\cdot \text{m}$$. The work done by the external agent is equal to:
A. $$4\times {10}^{-27}\text{J}$$
B. $$-4\times {10}^{-27}\text{J}$$
C. $$2.8\times {10}^{-26}\text{J}$$
D. $$-2.8\times {10}^{-26}\text{J}$$

Answer 1

Hint: Use the formula for the work done by an external agent on the electric dipole. Determine the initial and final work done by the external agent on the electric dipoles and then determine the net work done by subtracting the initial work done from the final work done by the external agent.

Formulae used:
The work done $$W$$ by an external agent on an electric dipole is given by
$$W=-\overrightarrow{p}\cdot \overrightarrow{E}$$ …… (1)
Here, $$\overrightarrow{p}$$ is the dipole moment and $$\overrightarrow{E}$$ is the electric field.

Complete step by step answer:
We can see from the given information that the initial electric dipole moment $$\overrightarrow{p}$$ is $$\overrightarrow{p}=\left(3\hat{i}+4\hat{j}\right)\times {10}^{-30}\text{C}\cdot \text{m}$$ and the final electric dipole moment $${\overrightarrow{p}}_f$$ is $$\left(-4\hat{i}+3\hat{j}\right)\times {10}^{-30}\text{C}\cdot \text{m}$$.
 $$\overrightarrow{p}=\left(3\hat{i}+4\hat{j}\right)\times {10}^{-30}\text{C}\cdot \text{m}$$
$${\overrightarrow{p}}_f=\left(-4\hat{i}+3\hat{j}\right)\times {10}^{-30}\text{C}\cdot \text{m}$$

The electric field is directed along X-direction and is given by $$\overrightarrow{E}=4000\hat{i}\text{N}\cdot \text{c}$$.

Let us determine the work done $$W_i$$ by the external agent on the initial electric dipole moment $$\overrightarrow{p}$$.
Rewrite equation (1) for the initial work done by the external agent on the electric dipole.
$$W_i=-\overrightarrow{p}\cdot \overrightarrow{E}$$

Substitute $$\overrightarrow{p}=\left(3\hat{i}+4\hat{j}\right)\times {10}^{-30}\text{C}\cdot \text{m}$$ for $$\overrightarrow{p}$$ and $$\overrightarrow{E}=4000\hat{i}\text{N}\cdot \text{c}$$ for $$\overrightarrow{E}$$ in the above equation.
$$W_i=-\left[\left(3\hat{i}+4\hat{j}\right)\times {10}^{-30}\text{C}\cdot \text{m}\right]\cdot \left(4000\hat{i}\text{N}\cdot \text{c}\right)$$
$$\Rightarrow W_i=-\left[\left(3\hat{i}\right)\times {10}^{-30}\text{C}\cdot \text{m}\right]\left(4000\hat{i}\text{N}\cdot \text{c}\right)-\left[\left(4\hat{j}\right)\times {10}^{-30}\text{C}\cdot \text{m}\right]\left(4000\hat{i}\text{N}\cdot \text{c}\right)$$
$$\Rightarrow W_i=-\left[\left(3\hat{i}\right)\times {10}^{-30}\text{C}\cdot \text{m}\right]\left(4000\hat{i}\text{N}\cdot \text{c}\right)-0$$
$$\Rightarrow W_i=-12000\times {10}^{-30}\text{J}$$
$$\Rightarrow W_i=-12\times {10}^{-27}\text{J}$$

Hence, the work done by the external agent on the initial electric dipole moment is $$-12\times {10}^{-27}\text{J}$$.

Let us determine the work done $$W_i$$ by the external agent on the initial electric dipole moment $$\overrightarrow{p}$$.
Rewrite equation (1) for the final work done $$W_f$$ by the external agent on the electric dipole.
$$W_i=-{\overrightarrow{p}}_f\cdot \overrightarrow{E}$$

Substitute $$\left(-4\hat{i}+3\hat{j}\right)\times {10}^{-30}\text{C}\cdot \text{m}$$ for $${\overrightarrow{p}}_f$$ and $$\overrightarrow{E}=4000\hat{i}\text{N}\cdot \text{c}$$ for $$\overrightarrow{E}$$ in the above equation.
$$W_f=-\left[\left(-4\hat{i}+3\hat{j}\right)\times {10}^{-30}\text{C}\cdot \text{m}\right]\cdot \left(4000\hat{i}\text{N}\cdot \text{c}\right)$$
$$\Rightarrow W_f=\left[\left(4\hat{i}\right)\times {10}^{-30}\text{C}\cdot \text{m}\right]\left(4000\hat{i}\text{N}\cdot \text{c}\right)-\left[\left(3\hat{j}\right)\times {10}^{-30}\text{C}\cdot \text{m}\right]\left(4000\hat{i}\text{N}\cdot \text{c}\right)$$
$$\Rightarrow W_f=\left[\left(4\hat{i}\right)\times {10}^{-30}\text{C}\cdot \text{m}\right]\left(4000\hat{i}\text{N}\cdot \text{c}\right)-0$$
$$\Rightarrow W_f=16000\times {10}^{-30}\text{J}$$
$$\Rightarrow W_f=16\times {10}^{-27}\text{J}$$

Hence, the work done by the external agent on the final electric dipole moment is $$16\times {10}^{-27}\text{J}$$.

The net work done $$W$$ by the external agent is
$$W=W_f-W_i$$

Substitute $$-12\times {10}^{-27}\text{J}$$ for $$W_i$$ and $$16\times {10}^{-27}\text{J}$$ for $$W_f$$ in the above equation.
$$W=\left(16\times {10}^{-27}\text{J}\right)-\left(-12\times {10}^{-27}\text{J}\right)$$
$$\Rightarrow W=2.8\times {10}^{-26}\text{J}$$

Therefore, the work done by the external agent is $$2.8\times {10}^{-26}\text{J}$$.

So, the correct answer is option (C).

Note:
While calculating the initial and final work done, the students may multiply all the electric dipole moment vectors by the electric field vector as a normal multiplication. But one should keep in mind that the dot product of two unit vectors along two different directions is zero and that of the two unit vectors along one direction is one.