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Chapter 11: Problem 1559

An electric dipole is placed at an angle of \(60^{\circ}\) with an electric field of intensity \(10^{5} \mathrm{NC}^{-1}\). It experiences a torque equal to \(8 \sqrt{3} \mathrm{Nm}\). If the dipole length is \(2 \mathrm{~cm}\) then the charge on the dipole is \(\ldots \ldots \ldots\) c. (A) \(-8 \times 10^{3}\) (B) \(8.54 \times 10^{-4}\) (C) \(8 \times 10^{-3}\) (D) \(0.85 \times 10^{-6}\)

Short Answer

Expert verified
The charge on the dipole is \(8 \times 10^{-3}\) C.

Step by step solution

01

Identify the given values

We are given the following values: 1. Angle between the electric dipole and the electric field θ = \(60^{\circ}\) 2. Electric field intensity E = \(10^{5} \mathrm{NC}^{-1}\) 3. Torque experienced by the dipole τ = \(8 \sqrt{3} \mathrm{Nm}\) 4. Length of the dipole d= \(2 \,\mathrm{cm = 0.02\,m}\) We need to find the charge, q, on the dipole.
02

Rewrite the formula for the torque

The torque experienced by a dipole in an electric field can be written as: τ = pEsinθ The dipole moment p is given by the product of the charge (q) and the distance (d) between the charges, so p = qd. Substitute this expression for the dipole moment into the torque formula: τ = qdEsinθ
03

Solve for the charge q

We need to solve for q in the equation: τ = qdEsinθ Divide both sides by dEsinθ to isolate q: q = \(\frac{\tau}{dE\sin\theta}\) Next, substitute the given values into the equation: q = \(\frac{8\sqrt{3}\,\mathrm{Nm}}{(0.02\,\mathrm{m})(10^{5}\,\mathrm{NC}^{-1})\sin{60^{\circ}}}\)
04

Calculate the charge q

We can now calculate the value of the charge on the dipole: q = \(\frac{8\sqrt{3}\,\mathrm{Nm}}{(0.02\,\mathrm{m})(10^{5}\,\mathrm{NC}^{-1})\left(\frac{\sqrt{3}}{2}\right)} = \frac{8\sqrt{3}\,\mathrm{Nm}}{0.01\,\mathrm{m}\cdot10^{5}\,\mathrm{NC}^{-1}\cdot\sqrt{3}}\) On simplification: q = \(8 \times 10^{-3}\, \mathrm{C}\) Comparing this value to the given choices, we find that the correct option is (C) \(8 \times 10^{-3}\) C.

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Most popular questions from this chapter

A parallel plate air capacitor has a capacitance \(18 \mu \mathrm{F}\). If the distance between the plates is tripled and a dielectric medium is introduced, the capacitance becomes \(72 \mu \mathrm{F}\). The dielectric constant of the medium is (A) 4 (B) 12 (C) 9 (D) 2

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