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Electrostatics question

2025 · 4 Apr · Shift 1 · Q70
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Electrostatics question

2025 · 4 Apr · Shift 1 · Q70

JEE MainPhysicsElectrostaticsMCQ+4 / −1
Two small spherical balls of mass 10 g each with charges −2μC-2 \mu \mathrm{C}−2μC and 2μC2 \mu \mathrm{C}2μC, are attached to two ends of very light rigid rod of length 20 cm . The arrangement is now placed near an infinite nonconducting charge sheet with uniform charge density of 100μC/m2100 \mu \mathrm{C} / \mathrm{m}^2100μC/m2 such that length of rod makes an angle of 30∘30^{\circ}30∘ with electric field generated by charge sheet. Net torque acting on the rod is: (Take εo:8.85×10−12C2/Nm2\varepsilon_{\mathrm{o}}: 8.85 \times 10^{-12} \mathrm{C}^2 / \mathrm{Nm}^2εo​:8.85×10−12C2/Nm2 )
  1. A
    1.12 Nm
  2. B
    2.24 Nm
  3. C
    11.2 Nm
  4. D
    112 Nm
View written solutionFree

Correct answer: A

  1. Electric field due to an infinite non-conducting sheet

For an infinite non-conducting sheet with uniform surface charge density σ\sigmaσ, the electric field is

E=σ2ε0E = \frac{\sigma}{2\varepsilon_0}E=2ε0​σ​

Given:

σ=100 μC/m2=100×10−6 C/m2=10−4 C/m2\sigma = 100\,\mu \text{C/m}^2 = 100 \times 10^{-6}\,\text{C/m}^2 = 10^{-4}\,\text{C/m}^2σ=100μC/m2=100×10−6C/m2=10−4C/m2

So,

E=10−42(8.85×10−12)E = \frac{10^{-4}}{2(8.85 \times 10^{-12})}E=2(8.85×10−12)10−4​ E≈10−41.77×10−11≈5.65×106 N/CE \approx \frac{10^{-4}}{1.77 \times 10^{-11}} \approx 5.65 \times 10^6\,\text{N/C}E≈1.77×10−1110−4​≈5.65×106N/C
  1. Forces on the two charges

Charges are +2 μC+2\,\mu C+2μC and −2 μC-2\,\mu C−2μC.

Magnitude of force on each charge:

F=qE=2×10−6×5.65×106F = qE = 2 \times 10^{-6} \times 5.65 \times 10^6F=qE=2×10−6×5.65×106 F≈11.3 NF \approx 11.3\,\text{N}F≈11.3N

The positive charge experiences force along the field, and the negative charge opposite to the field. These two equal and opposite forces form a couple.

  1. Torque due to the couple

Torque of a couple is

τ=F×d\tau = F \times dτ=F×d

where ddd is the perpendicular distance between the two parallel lines of action of the forces.

The rod length is

L=20 cm=0.2 mL = 20\,\text{cm} = 0.2\,\text{m}L=20cm=0.2m

The rod makes angle 30∘30^\circ30∘ with the electric field, so perpendicular separation between the forces is

d=Lsin⁡30∘=0.2×12=0.1 md = L\sin 30^\circ = 0.2 \times \frac{1}{2} = 0.1\,\text{m}d=Lsin30∘=0.2×21​=0.1m

Hence,

τ=11.3×0.1=1.13 N m\tau = 11.3 \times 0.1 = 1.13\,\text{N m}τ=11.3×0.1=1.13N m

Approximately,

τ≈1.12 N m\tau \approx 1.12\,\text{N m}τ≈1.12N m
  1. Check using dipole formula

Dipole moment magnitude:

p=qL=2×10−6×0.2=4×10−7 C mp = qL = 2 \times 10^{-6} \times 0.2 = 4 \times 10^{-7}\,\text{C m}p=qL=2×10−6×0.2=4×10−7C m

Torque on dipole:

τ=pEsin⁡θ\tau = pE\sin\thetaτ=pEsinθ

with θ=30∘\theta = 30^\circθ=30∘:

τ=4×10−7×5.65×106×12\tau = 4 \times 10^{-7} \times 5.65 \times 10^6 \times \frac{1}{2}τ=4×10−7×5.65×106×21​ τ≈1.13 N m\tau \approx 1.13\,\text{N m}τ≈1.13N m

Same result.

  1. Final answer
τ≈1.12 N m\boxed{\tau \approx 1.12\,\text{N m}}τ≈1.12N m​

So the correct option is A.

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