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Physics Question on Electric charges and fields

An electric dipole consists of two opposite charges, each of magnitude 1.0 μC1.0\, \mu C separated by a distance of 2.0 cm2.0 \, cm. The dipole is placed in an external field of 105 NC−110^5 \, NC^{-1}. The maximum torque on the dipole is

A

\ce0.2×10−3 Nm \ce{ 0.2 \times 10^{-3} \, N m}

B

\ce1×10−3 Nm \ce{ 1 \times 10^{-3} \, N m}

C

\ce2×10−3 Nm \ce{ 2 \times 10^{-3} \, N m}

D

\ce4×10−3 Nm \ce{ 4 \times 10^{-3} \, N m}

Answer

\ce4×10−3 Nm \ce{ 4 \times 10^{-3} \, N m}

Explanation

Solution

1×10−6C1 \times 10^{-6} C distance =2cm=2 cm Exter field =105N/C=10^{5} N / C maximum torque on the dipole == ? q=1×10−6C,2a=2cm  or, =0.02cm ∴P=q×2a =(1×10−6)×0.02 =2×10−8cm\begin{array}{l} q =1 \times 10^{-6} C , \quad 2 a =2 cm \\\ \text { or, }=0.02 cm \\\ \therefore \quad P & = q \times 2 a \\\ & =\left(1 \times 10^{-6}\right) \times 0.02 \\\ & =2 \times 10^{-8} cm \end{array} Intensity of the external electric field, E=1.0×105N/CE =1.0 \times 10^{5} N / C
(i) Zmax⁡=pE=(2×10−8)(10×105)=2×10−3N−mZ_{\max }=p E=\left(2 \times 10^{-8}\right)\left(10 \times 10^{5}\right)=2 \times 10^{-3} N - m
(ii) Net work done in turning the dipole from 000^{0} to 180∘180^{\circ}
i.e W=∫0∘180∘r‾dθ=∫0∘180∘pEsin⁡θdθW =\int_{0^{\circ}}^{180^{\circ}} \overline{ r } d \theta=\int_{0^{\circ}}^{180^{\circ}} pE \sin \theta d \theta

=pE[−cos⁡θ]00180∘ =−pE(cos⁡180∘−cos⁡00) =2pE =2×(2×10−8)(1×105)J =4×10−3J\begin{array}{l} = pE [-\cos \theta]_{0^{0}}^{180^{\circ}} \\\ =- pE \left(\cos 180^{\circ}-\cos 0^{0}\right) \\\ =2 pE \\\ =2 \times\left(2 \times 10^{-8}\right)\left(1 \times 10^{5}\right) J \\\ =4 \times 10^{-3} J \end{array}