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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

Two metal plate form a parallel plate capacitor. The distance between the plates is \(\mathrm{d}\). A metal sheet of thickness \(\mathrm{d} / 2\) and of the same area is introduced between the plates. What is the ratio of the capacitance in the two cases? (A) \(4: 1\) (B) \(3: 1\) (C) \(2: 1\) (D) \(5: 1\)

Two uniformaly charged spherical conductors \(\mathrm{A}\) and \(\mathrm{B}\) having radius \(1 \mathrm{~mm}\) and \(2 \mathrm{~mm}\) are separated by a distance of \(5 \mathrm{~cm} .\) If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres \(\mathrm{A}\) and \(\mathrm{B}\) is..... (A) \(4: 1\) (B) \(1: 2\) (C) \(2: 1\) (D) \(1: 4\)

Three charges, each of value \(Q\), are placed at the vertex of an equilateral triangle. A fourth charge \(q\) is placed at the centre of the triangle. If the charges remains stationery then, \(q=\ldots \ldots \ldots\) (A) \((\mathrm{Q} / \sqrt{2})\) (B) \(-(\mathrm{Q} / \sqrt{3})\) (C) \(-(Q / \sqrt{2})\) (D) \((\mathrm{Q} / \sqrt{3})\)

Electric potential at any point is $\mathrm{V}=-5 \mathrm{x}+3 \mathrm{y}+\sqrt{(15 \mathrm{z})}$, then the magnitude of the electric field is \(\ldots \ldots \ldots \mathrm{N} / \mathrm{C}\). (A) \(3 \sqrt{2}\) (B) \(4 \sqrt{2}\) (C) 7 (D) \(5 \sqrt{2}\)

Two point charges \(100 \mu \mathrm{c}\) and \(5 \mu \mathrm{c}\) are placed at points \(\mathrm{A}\) and \(B\) respectively with \(A B=40 \mathrm{~cm}\). The work done by external force in displacing the charge \(5 \mu \mathrm{c}\) from \(\mathrm{B}\) to \(\mathrm{C}\) where \(\mathrm{BC}=30 \mathrm{~cm}\), angle \(\mathrm{ABC}=(\pi / 2)\) and \(\left[1 /\left(4 \pi \epsilon_{0}\right)\right]\) \(=9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{c}^{2}\). (A) \(9 \mathrm{~J}\) (B) \((9 / 25) \mathrm{J}\) (C) \((81 / 20) \mathrm{J}\) (D) \(-(9 / 4) \mathrm{J}\)
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