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Chapter 11: Problem 1536

Three identical spheres each having a charge \(\mathrm{q}\) and radius \(R\), are kept in such a way that each touches the other two spheres. The magnitude of the electric force on any sphere due to other two is \(\ldots \ldots \ldots\) (A) $(\mathrm{R} / 2)\left[1 /\left(4 \pi \epsilon_{0}\right)\right](\sqrt{5} / 4)(\mathrm{q} / \mathrm{R})^{2}$ (B) $\left[1 /\left(8 \pi \epsilon_{0}\right)\right](\sqrt{2} / 3)(\mathrm{q} / \mathrm{R})^{2}$ (C) $\left[1 /\left(4 \pi \epsilon_{0}\right)\right](\sqrt{3} / 4)(\mathrm{q} / \mathrm{R})^{2}$ (D) $-\left[1 /\left(8 \pi \epsilon_{0}\right)\right](\sqrt{3} / 2)(\mathrm{q} / \mathrm{R})^{2}$

Short Answer

Expert verified
The magnitude of the electric force on any sphere due to the other two spheres is \(\left[1 /\left(4 \pi \epsilon_{0}\right)\right](\sqrt{3} / 4)(q / R)^{2}\).

Step by step solution

01

Identifying Information

From the given problem, we know that each sphere has a charge q and radius R.
02

Determine the Distance Between the Charges on Each Sphere

Since the spheres touch each other, the distance between the centers of any two spheres is the sum of their radii. As their radii are equal (\(R\)), the distance between the charges on two spheres is: \(d = 2R\)
03

Calculate Electric Force Between One Pair of Spheres

Using Coulomb's Law, find the electric force (\(F_{12}\)) between two spheres. Coulomb's Law states that the force between two charges is given by: \(F=k \cdot \frac{q_1 \cdot q_2}{r^2}\), where \(k=\frac{1}{4 \pi \epsilon_0}\) Since both charges are equal (q), the force on any one sphere (1) due to the other sphere (2) is: \(F_{12}=\left(\frac{1}{4 \pi \epsilon_{0}}\right)\frac{q^2}{(2R)^2}\)
04

Calculate the Net Electric Force on One Sphere

Given three spheres forming a triangle, let us label the spheres A, B, and C. To find the net electric force on sphere A, we need to sum the electric forces acting on it, i.e., forces due to sphere B and sphere C. First, find the angle \(\theta\) between \(F_{AB}\) and \(F_{AC}\). Since they form an equilateral triangle, \(\theta= 60^{\circ}\) Now, calculate the net force by adding forces \(F_{AB}\) and \(F_{AC}\) in a vector form. Considering that both forces have a magnitude of \(F_{12}\), the net force can be determined using the cosine rule as: \(F_A = \sqrt{{F_{12}}^2 + {F_{12}}^2 + 2 \cdot F_{12} \cdot F_{12} \cdot \cos{60^{\circ}}}\)
05

Simplify the Net Electric Force Formula

Substitute \(F_{12}\) and \(\cos{60^{\circ}} = \frac{1}{2}\) in the above equation and simplify: \(F_A = \sqrt{{\left(\frac{1}{4 \pi \epsilon_{0}}\frac{q^2}{(2R)^2}\right)}^2 + {\left(\frac{1}{4 \pi \epsilon_{0}}\frac{q^2}{(2R)^2}\right)}^2 + 2 \cdot \left(\frac{1}{4 \pi \epsilon_{0}}\frac{q^2}{(2R)^2}\right) \cdot \left(\frac{1}{4 \pi \epsilon_{0}}\frac{q^2}{(2R)^2}\right) \cdot \frac{1}{2}}\) After simplifying, we obtain: \(F_A = \left[1 /\left(4 \pi \epsilon_{0}\right)\right](\sqrt{3} / 4)(q / R)^{2}\) Comparing with the given options, the magnitude of the electric force on any sphere due to the other two spheres is answer choice (C).

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Most popular questions from this chapter

Two Points \(P\) and \(Q\) are maintained at the Potentials of \(10 \mathrm{v}\) and \(-4 \mathrm{v}\), respectively. The work done in moving 100 electrons from \(\mathrm{P}\) to \(\mathrm{Q}\) is \(\ldots \ldots \ldots\) (A) \(2.24 \times 10^{-16} \mathrm{~J}\) (B) \(-9.60 \times 10^{-17} \mathrm{~J}\) (C) \(-2.24 \times 10^{-16} \mathrm{~J}\) (D) \(9.60 \times 10^{-17} \mathrm{~J}\)

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An electric dipole is placed along the \(\mathrm{x}\) -axis at the origin o. \(\mathrm{A}\) point \(P\) is at a distance of \(20 \mathrm{~cm}\) from this origin such that OP makes an angle \((\pi / 3)\) with the x-axis. If the electric field at P makes an angle \(\theta\) with the x-axis, the value of \(\theta\) would be \(\ldots \ldots \ldots\) (A) \((\pi / 3)+\tan ^{-1}(\sqrt{3} / 2)\) (B) \((\pi / 3)\) (C) \((2 \pi / 3)\) (D) \(\tan ^{-1}(\sqrt{3} / 2)\)
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