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JEE Main 202313 AprilMorning Shift
Exam
JEE Main
Year
2023
Date
13 April
Shift
Morning Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2023 · 13 April · Morning Shift · Physics · Electrostatics

Two charges of each magnitude 0 . 01C and separated by a distance...

Two charges of each magnitude 0.01C0 . 01C and separated by a distance of 0.4mm0 . 4mm constitute an electric dipole. If the dipole is placed in an uniform electric field E→\overset{\to}{E} of 10dyne⋅C−110dyne \cdot {C}^{- 1} making 30∘{30}^{∘} angle with E→,\overset{\to}{E}, the magnitude of torque acting on dipole is:

Options

  1. A4.0×10−10Nm4 . 0 \times {10}^{- 10}Nm
  2. B1.0×10−8Nm1 . 0 \times {10}^{- 8}Nm
  3. C1.5×10−9Nm1 . 5 \times {10}^{- 9}Nm
  4. D2.0×10−10Nm2 . 0 \times {10}^{- 10}Nm

Correct answer

D. 2.0×10−10Nm2 . 0 \times {10}^{- 10}Nm

Step-by-step solution

The formula to calculate the dipole moment is given by

∣p⇀∣=q∣d⇀∣...(1)\left|\overset{⇀}{p}\right| = q\left|\overset{⇀}{d}\right| . . . \left(1\right)

Substitute the values of the known parameters into equation (1) to calculate the dipole moment.

{∣p⇀∣=0.01C×0.4mm×1m103mm=4.0×10−6C⋅m\begin{cases}\left|\overset{⇀}{p}\right| & = & 0 . 01C \times 0 . 4mm \times \dfrac{1m}{{10}^{3}mm} \\ & = & 4 . 0 \times {10}^{- 6}C \cdot m\end{cases}

The formula to calculate the torque on the dipole in the electric field is given by

∣τ→∣=∣p→×E→∣...(2)| \overset{\to}{τ} | = | \overset{\to}{p} \times \overset{\to}{E} | . . . \left(2\right)

Substitute the values of the known parameters into equation (2) to calculate the required torque.

{∣τ⇀∣=4.0×10−6C⋅m×10dyne⋅C−1×1N105dyne×sin⁡30°=2×10−10Nm\begin{cases}\left|\overset{⇀}{τ}\right| & = & 4 . 0 \times {10}^{- 6}C \cdot m \times 10dyne \cdot {C}^{- 1} \times \dfrac{1N}{{10}^{5}dyne} \times \sin 30 ° \\ & = & 2 \times {10}^{- 10}Nm\end{cases}

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