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Electrostatics question

2019 · 11 Jan · Shift 2 · Q65
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Electrostatics question

2019 · 11 Jan · Shift 2 · Q65

JEE MainPhysicsElectrostaticsMCQ+4 / −1
An electric field of 1000 V/m is applied to an electric dipole at angle of 45o. The value of electric dipole moment is 10–29 C.m. What is the potential energy of the electric dipole?
  1. A
    - 7 ×\times× 10–27 J
  2. B
    −-− 9 ×\times× 10–20 J
  3. C
    −-− 10 ×\times× 10–29 J
  4. D
    −-− 20 ×\times× 10–18 J
View written solutionFree

Correct answer: A

  1. Use the formula for potential energy of a dipole in a uniform electric field

    U=−p⃗⋅E⃗=−pEcos⁡θU = -\vec{p}\cdot\vec{E} = -pE\cos\thetaU=−p​⋅E=−pEcosθ

    where:

    • p=10−29 C⋅mp = 10^{-29}\ \text{C·m}p=10−29 C⋅m
    • E=1000 V/m=103 V/mE = 1000\ \text{V/m} = 10^3\ \text{V/m}E=1000 V/m=103 V/m
    • θ=45∘\theta = 45^\circθ=45∘

  2. Substitute the values

    U=−(10−29)(103)cos⁡45∘U = -(10^{-29})(10^3)\cos 45^\circU=−(10−29)(103)cos45∘

  3. Use cos⁡45∘=12≈0.707\cos 45^\circ = \frac{1}{\sqrt{2}} \approx 0.707cos45∘=2​1​≈0.707

    U=−10−26×0.707U = -10^{-26} \times 0.707U=−10−26×0.707

    U≈−7.07×10−27 JU \approx -7.07 \times 10^{-27}\ \text{J}U≈−7.07×10−27 J

  4. Match with the given options

    U≈−7×10−27 JU \approx -7 \times 10^{-27}\ \text{J}U≈−7×10−27 J

    So the correct option is A.

  5. Option check

    • A: −7×10−27 J-7 \times 10^{-27}\,\text{J}−7×10−27J ✅ matches
    • B: −9×10−20 J-9 \times 10^{-20}\,\text{J}−9×10−20J ❌ wrong power
    • C: −10×10−29 J=−10−28 J-10 \times 10^{-29}\,\text{J} = -10^{-28}\,\text{J}−10×10−29J=−10−28J ❌ too small
    • D: −20×10−18 J-20 \times 10^{-18}\,\text{J}−20×10−18J ❌ wrong power

Therefore, the potential energy is:

−7×10−27 J\boxed{-7 \times 10^{-27}\ \text{J}}−7×10−27 J​

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