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Question: An electric field of 1000 V/m is applied to an electric dipole at angle of 45°. The value of electri...

An electric field of 1000 V/m is applied to an electric dipole at angle of 45°. The value of electric dipole moment is 10−2910^{-29} C.m. What is the potential energy of the electric dipole?

A
  • 7 × 10−2710^{-27} J
B
  • 20 × 10−1810^{-18} J
C
  • 9 × 10−2010^{-20} J
D
  • 10 × 10−2910^{-29} J
Answer
  • 7 × 10−2710^{-27} J
Explanation

Solution

The potential energy UU of an electric dipole with dipole moment p\mathbf{p} in a uniform electric field E\mathbf{E} is given by the formula: U=−p⋅EU = - \mathbf{p} \cdot \mathbf{E}

In scalar form, this is written as: U=−pEcos⁡θU = - p E \cos \theta

where pp is the magnitude of the electric dipole moment, EE is the magnitude of the electric field, and θ\theta is the angle between the electric dipole moment vector p\mathbf{p} and the electric field vector E\mathbf{E}.

Given values are: Electric field strength, E=1000E = 1000 V/m =103= 10^3 V/m Electric dipole moment, p=10−29p = 10^{-29} C.m Angle between p\mathbf{p} and E\mathbf{E}, θ=45∘\theta = 45^\circ

Substitute these values into the formula for potential energy: U=−(10−29 C.m)×(103 V/m)×cos⁡(45∘)U = - (10^{-29} \text{ C.m}) \times (10^3 \text{ V/m}) \times \cos(45^\circ) U=−(10−29×103)×cos⁡(45∘)U = - (10^{-29} \times 10^3) \times \cos(45^\circ) J U=−10−26×cos⁡(45∘)U = - 10^{-26} \times \cos(45^\circ) J

The value of cos⁡(45∘)\cos(45^\circ) is 12\frac{1}{\sqrt{2}}. U=−10−26×12U = - 10^{-26} \times \frac{1}{\sqrt{2}} J U=−12×10−26U = - \frac{1}{\sqrt{2}} \times 10^{-26} J

To compare with the given options, we can approximate 12\frac{1}{\sqrt{2}}. 12≈0.707\frac{1}{\sqrt{2}} \approx 0.707 U≈−0.707×10−26U \approx - 0.707 \times 10^{-26} J U≈−7.07×10−27U \approx - 7.07 \times 10^{-27} J

The calculated value −7.07×10−27-7.07 \times 10^{-27} J is very close to −7×10−27-7 \times 10^{-27} J.