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Electrostatics question

2024 · 27 Jan · Shift 2 · Q83
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Electrostatics question

2024 · 27 Jan · Shift 2 · Q83

JEE MainPhysicsElectrostaticsNumerical+4 / −1
Two charges of −4μC-4 \mu \mathrm{C}−4μC and +4μC+4 \mu \mathrm{C}+4μC are placed at the points A(1,0,4)m\mathrm{A}(1,0,4) \mathrm{m}A(1,0,4)m and B(2,−1,5)m\mathrm{B}(2,-1,5) \mathrm{m}B(2,−1,5)m located in an electric field E→=0.20i^ V/cm\overrightarrow{\mathrm{E}}=0.20 \hat{i} \mathrm{~V} / \mathrm{cm}E=0.20i^ V/cm. The magnitude of the torque acting on the dipole is 8α×10−5Nm8 \sqrt{\alpha} \times 10^{-5} \mathrm{Nm}8α​×10−5Nm, where α=‾\alpha=\underline{\hspace{2cm}}α=​.
Numerical answer
View written solutionFree

Correct answer: 2

  1. Identify the dipole charges and positions

The charges are:

  • −4 μC-4\,\mu C−4μC at A(1,0,4)A(1,0,4)A(1,0,4) m
  • +4 μC+4\,\mu C+4μC at B(2,−1,5)B(2,-1,5)B(2,−1,5) m

So the dipole moment vector is from negative to positive charge, i.e. from AAA to BBB.

AB⃗=(2−1)i^+(−1−0)j^+(5−4)k^=i^−j^+k^\vec{AB} = (2-1)\hat i + (-1-0)\hat j + (5-4)\hat k = \hat i - \hat j + \hat kAB=(2−1)i^+(−1−0)j^​+(5−4)k^=i^−j^​+k^

Hence,

p⃗=q AB⃗=4×10−6(i^−j^+k^)  C m\vec p = q\,\vec{AB} = 4\times 10^{-6}(\hat i - \hat j + \hat k)\;\text{C m}p​=qAB=4×10−6(i^−j^​+k^)C m
  1. Write the electric field in SI units

Given:

E⃗=0.20 i^  V/cm\vec E = 0.20\,\hat i\;\text{V/cm}E=0.20i^V/cm

Since 1 V/cm=100 V/m1\,\text{V/cm} = 100\,\text{V/m}1V/cm=100V/m,

E⃗=0.20×100 i^=20 i^  V/m\vec E = 0.20\times 100\,\hat i = 20\,\hat i\;\text{V/m}E=0.20×100i^=20i^V/m
  1. Use torque formula

The torque on an electric dipole in a uniform electric field is

τ⃗=p⃗×E⃗\vec \tau = \vec p \times \vec Eτ=p​×E

Its magnitude is

τ=pEsin⁡θ\tau = pE\sin\thetaτ=pEsinθ

Since E⃗\vec EE is along i^\hat ii^, only the component of p⃗\vec pp​ perpendicular to i^\hat ii^ contributes.

From

p⃗=4×10−6(i^−j^+k^)\vec p = 4\times 10^{-6}(\hat i - \hat j + \hat k)p​=4×10−6(i^−j^​+k^)

perpendicular part is

p⃗⊥=4×10−6(−j^+k^)\vec p_\perp = 4\times 10^{-6}(-\hat j + \hat k)p​⊥​=4×10−6(−j^​+k^)

So,

∣p⃗⊥∣=4×10−6(−1)2+12=4×10−62|\vec p_\perp| = 4\times 10^{-6}\sqrt{(-1)^2 + 1^2} = 4\times 10^{-6}\sqrt{2}∣p​⊥​∣=4×10−6(−1)2+12​=4×10−62​

Therefore,

τ=E ∣p⃗⊥∣=20×4×10−62\tau = E\,|\vec p_\perp| = 20\times 4\times 10^{-6}\sqrt{2}τ=E∣p​⊥​∣=20×4×10−62​ τ=80×10−62=82×10−5  N m\tau = 80\times 10^{-6}\sqrt{2} = 8\sqrt{2}\times 10^{-5}\;\text{N m}τ=80×10−62​=82​×10−5N m
  1. Compare with the given form

Given,

τ=8α×10−5  N m\tau = 8\sqrt{\alpha}\times 10^{-5}\;\text{N m}τ=8α​×10−5N m

Matching,

8α×10−5=82×10−58\sqrt{\alpha}\times 10^{-5} = 8\sqrt{2}\times 10^{-5}8α​×10−5=82​×10−5

Hence,

α=2\alpha = 2α=2
  1. Comparison with stored answer

Stored correct answer = 222

Our derived answer also gives 222, so they agree.

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