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Question: An electric dipole with dipole moment \[\vec p = \left( {3\hat i + 4\hat j} \right) \times {10^{ - 3...

An electric dipole with dipole moment p⃗=(3i^+4j^)×10−30 C⋅m\vec p = \left( {3\hat i + 4\hat j} \right) \times {10^{ - 30}}\,{\text{C}} \cdot {\text{m}} is placed in an electric field E⃗=4000i^ N⋅c\vec E = 4000\hat i\,{\text{N}} \cdot {\text{c}}. An external agent turns the dipole and is dipole moment becomes (−4i^+3j^)×10−30 C⋅m\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. 4×10−27 J4 \times {10^{ - 27}}\,{\text{J}}
B. −4×10−27 J - 4 \times {10^{ - 27}}\,{\text{J}}
C. 2.8×10−26 J2.8 \times {10^{ - 26}}\,{\text{J}}
D. −2.8×10−26 J - 2.8 \times {10^{ - 26}}\,{\text{J}}

Explanation

Solution

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 WW by an external agent on an electric dipole is given by
W=−p⃗⋅E⃗W = - \vec p \cdot \vec E …… (1)
Here, p⃗\vec p is the dipole moment and E⃗\vec E is the electric field.

Complete step by step answer:
We can see from the given information that the initial electric dipole moment p⃗\vec p is p⃗=(3i^+4j^)×10−30 C⋅m\vec p = \left( {3\hat i + 4\hat j} \right) \times {10^{ - 30}}\,{\text{C}} \cdot {\text{m}} and the final electric dipole moment p⃗f{\vec p_f} is (−4i^+3j^)×10−30 C⋅m\left( { - 4\hat i + 3\hat j} \right) \times {10^{ - 30}}\,{\text{C}} \cdot {\text{m}}.
p⃗=(3i^+4j^)×10−30 C⋅m\vec p = \left( {3\hat i + 4\hat j} \right) \times {10^{ - 30}}\,{\text{C}} \cdot {\text{m}}
p⃗f=(−4i^+3j^)×10−30 C⋅m{\vec 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 E⃗=4000i^ N⋅c\vec E = 4000\hat i\,{\text{N}} \cdot {\text{c}}.

Let us determine the work done Wi{W_i} by the external agent on the initial electric dipole moment p⃗\vec p.
Rewrite equation (1) for the initial work done by the external agent on the electric dipole.
Wi=−p⃗⋅E⃗{W_i} = - \vec p \cdot \vec E

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

Let us determine the work done Wi{W_i} by the external agent on the initial electric dipole moment p⃗\vec p.
Rewrite equation (1) for the final work done Wf{W_f} by the external agent on the electric dipole.
Wi=−p⃗f⋅E⃗{W_i} = - {\vec p_f} \cdot \vec E

Substitute (−4i^+3j^)×10−30 C⋅m\left( { - 4\hat i + 3\hat j} \right) \times {10^{ - 30}}\,{\text{C}} \cdot {\text{m}} for p⃗f{\vec p_f} and E⃗=4000i^ N⋅c\vec E = 4000\hat i\,{\text{N}} \cdot {\text{c}} for E⃗\vec E in the above equation.
Wf=−[(−4i^+3j^)×10−30 C⋅m]⋅(4000i^ N⋅c){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)
⇒Wf=[(4i^)×10−30 C⋅m](4000i^ N⋅c)−[(3j^)×10−30 C⋅m](4000i^ N⋅c)\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)
⇒Wf=[(4i^)×10−30 C⋅m](4000i^ N⋅c)−0\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
⇒Wf=16000×10−30 J\Rightarrow {W_f} = 16000 \times {10^{ - 30}}\,{\text{J}}
⇒Wf=16×10−27 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 16×10−27 J16 \times {10^{ - 27}}\,{\text{J}}.

The net work done WW by the external agent is
W=Wf−WiW = {W_f} - {W_i}

Substitute −12×10−27 J - 12 \times {10^{ - 27}}\,{\text{J}} for Wi{W_i} and 16×10−27 J16 \times {10^{ - 27}}\,{\text{J}} for Wf{W_f} in the above equation.
W=(16×10−27 J)−(−12×10−27 J)W = \left( {16 \times {{10}^{ - 27}}\,{\text{J}}} \right) - \left( { - 12 \times {{10}^{ - 27}}\,{\text{J}}} \right)
⇒W=2.8×10−26 J\Rightarrow W = 2.8 \times {10^{ - 26}}\,{\text{J}}

Therefore, the work done by the external agent is 2.8×10−26 J2.8 \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.