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Question: An electric dipole of length 20 cm having \(\pm 3 \times 10^{- 3}C\) charge placed at \(60^{o}\)with...

An electric dipole of length 20 cm having ±3×10−3C\pm 3 \times 10^{- 3}C charge placed at 60o60^{o}with respect to a uniform electric field experiences a torque of magnitude 6 N m. the potential energy of the dipole is.

A

−23J- 2\sqrt{3}J

B

53J5\sqrt{3}J

C

−32J- 3\sqrt{2}J

D

35J3\sqrt{5}J

Answer

−23J- 2\sqrt{3}J

Explanation

Solution

: Here, length of dipole, 2a=20cm=20×10−2m2a = 20cm = 20 \times 10^{- 2}m Charge)q=±3×10−3C,θ=60oq = \pm 3 \times 10^{- 3}C,\theta = 60^{o}and torque

τ=6NmAsτ=pEsin⁡θ\tau = 6NmAs\tau = pE\sin\theta

Or E=τpsin⁡θ=τq(2q)sin⁡θE = \frac{\tau}{p\sin\theta} = \frac{\tau}{q(2q)\sin\theta} (∵\becausep=q(2a))

∴\thereforePotential energy of dipole,

U=−pEcos⁡θ=−q(2a)Ecos⁡θU = - pE\cos\theta = - q(2a)E\cos\theta

=−3×10−3(20×10−2)10553cos⁡60o= - 3 \times 10^{- 3}(20 \times 10^{- 2})\frac{10^{5}}{5\sqrt{3}}\cos 60^{o}

=−3×10−5×20×10553×2=−23J= \frac{- 3 \times 10^{- 5} \times 20 \times 10^{5}}{5\sqrt{3} \times 2} = - 2\sqrt{3}J