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Physics Question on electrostatic potential and capacitance

An electric dipole of length 20 cm20 \,cm having ±3×10−3 C\pm 3 \times10^{-3}\,C charge placed at 60∘60^{\circ} with respect to a uniform electric field experiences a torque of magnitude 6 N m6\, N\, m. The potential energy of the dipole is

A

−23 J-2 \sqrt{3}\, J

B

53 J5\sqrt{3}\,J

C

−32 J-3 \sqrt{2}\,J

D

35 J3 \sqrt{5}\, J

Answer

−23 J-2 \sqrt{3}\, J

Explanation

Solution

Here, length of dipole, 2a=20 cm=20×10−2 m2a = 20 \,cm = 20 \times 10^{-2}\, m Charge q=± 3×10−3 C,θ=60∘q =\pm\, 3\times10^{-3}\, C, \theta=60^{\circ} and torque τ=6 N m\tau=6\,N\,m As τ=PEsinθ\tau=PEsin \theta or E=τP sin θ=τq(2a)sin θ(∵p=q(2a))E=\frac{\tau}{P\,sin\,\theta}=\frac{\tau}{q\left(2a\right)sin\,\theta} \left(\because p=q\left(2a\right)\right) ∴E=63×10−3×20×10−2×sin 60∘=10553N C−1\therefore E=\frac{6}{3\times10^{-3}\times20\times10^{-2}\times sin\,60^{\circ}}=\frac{10^{5}}{5\sqrt{3}}N \,C^{-1} Potential energy of dipole, U=−pEcosθ=−q(2a)EcosθU =-pEcos\theta=-q\left(2a\right)Ecos\theta =−3×10−3(20×10−2)10553cos 60∘=-3\times10^{-3}\left(20\times10^{-2}\right)\frac{10^{5}}{5\sqrt{3}}cos\,60^{\circ} =−3×10−5×20×10553×2=\frac{-3\times10^{-5}\times20\times10^{5}}{5\sqrt{3}\times2} =−23 J=-2\sqrt{3}\,J