An electric dipole is placed at an angle of $60^{\circ}$ with an electric filed of intensity $10^{5} {NC}^{- 1} .$ It experiences a torque equal to $8 \sqrt{3}$ Nm. Calculate the charge on dipole separated by 2 cm.

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$8 \times 10^{- 6} C$

$8 .54 \times 10^{- 4} C$

${8 × 10}^{- 3} C$

$0 .85 \times 10^{- 6} C$

answer is C.

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Detailed Solution

Here, $θ = 60^{\circ}, E = 10^{5} {NC}^{- 1}$

$τ = 8 \sqrt{3} Nm, q = ?, 2 a = 2 cm = 2 \times 10^{- 2} m$

From $τ = pE \sin θ = q (2 a) E \sin θ$

$q = \frac{τ}{2 a E \sin θ} = \frac{8 \sqrt{3}}{2 \times 10^{- 2} \times 10^{5} \times \sin 60^{\circ}}$

$= \frac{8 \sqrt{3}}{2 \times 10^{3} \times \frac{\sqrt{3}}{2}}$

$q = 8 \times 10^{- 3} C$

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