Question

An electric dipole of dipole moment 2x 10-12 C-m is oriented at an angle of 60⁰ with electric field of intensity 3NC¹. What work need to be done to turn the dipole so that it becomes anti-parallel to the field?

a. 12 x 10-12 J

c. 9 × 10-12J

b. -9 x 10-123

d. -12 x 10-12J​

Answer 1

Answer:

Electric dipole moment = P = 2×10^{-12} Cm

∅1 = 60°

Electric field = E = 3 N/C

If we turn the dipole antiparallel to the field, the angle between the field and the dipole will become 180°, an unstable condition.

Hence, ∅2 = -180° (Antiparallel)

Now, We know that, U = -PE(cos∅2-cos∅1)

Putting the given values,

U = -2×10^{-12}×3(cos(-180°)-cos60°)

Here, applying : cos(-x)=cos x.

⇒ U = -2×10^{-12}×3(cos180°-cos60°)

⇒ U = -6×10^{-12}×(-1-½)

⇒ U = -6×10^{-12}×{(-2-1)/2}

⇒ U = -6×10^{-12}×-3/2

⇒ U = -3×-3×10^{-12}

⇒ U = 9×10^{-12}

Hence, 9×10^{-12} J

Extra Information:

• Electric field intensity at any point due to source charge cam be defined as the force experienced per unit positive charge placed at a point without disturbing the source charge.
• E = F/q
• It's a vector quantity.
• SI unit : N/C
• Electric field of the positive charge is directed radially outward whereas the electric field of the negative charge is directed radially inward.
• Electric dipole is the system of two equal and opposite charges kept at a certain separation.
• Electric dipole moment is the strength of the electric dipole.
• P = 2aq
• It's a vector quantity.
• SI unit : C-m
• Direction of the electric dipole moment is from negative charge to the positive charge.

Answer 2

Given :-

An electric dipole of dipole moment 2x 10⁻¹² C-m is oriented at an angle of 60⁰ with electric field of intensity 3NC¹

To Find :-

Work need

Solution :-

We know that

$\sf U = -PE(\cos\theta_2 - \cos\theta_1)$

$\sf U = -\{2\times 10^{-12} \times 3 (\cos(-180) - \cos(60)\}$

$\sf U = -\{2 \times 3 \times 10^{-12}(-1 - 0.5)\}$

$\sf U = -\{6 \times 10^{-12} \times -1.5\}$

$\sf U = 9 \times 10^{-12}\;J$

Hence

Option C is correct