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Chapter 11: Problem 1577

Three particles, each having a charge of \(10 \mu \mathrm{c}\) are placed at the corners of an equilateral triangle of side \(10 \mathrm{~cm}\). The electrostatic potential energy of the system is (Given $\left.\left[1 /\left(4 \pi \epsilon_{0}\right)\right]=9 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{c}^{2}\right)$ (A) \(100 \mathrm{~J}\) (B) \(27 \mathrm{~J}\) (C) Zero (D) Infinite

Short Answer

Expert verified
The total electrostatic potential energy of the system is (B) \(27 J\).

Step by step solution

01

Define the given information

We are given the following information: - Charge of each particle: \(q = 10 \mu C = 10 \times 10^{-6} C\) - Side length of the equilateral triangle: \(d = 10 cm = 0.1 m\) - Coulomb's constant: \(\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 Nm^2/C^2\)
02

Calculate the electrostatic potential energy between each pair of particles

The formula for the electrostatic potential energy (U) between two particles with charges \(q_1\) and \(q_2\) separated by a distance r is: \[U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r}\] In our case, all three particles have the same charge, so \(q_1 = q_2 = q\), and they are separated by the same distance (the side length of the equilateral triangle), so \(r = d\). By applying the formula, we get the potential energy between each pair of particles: \[U_{pair} = \frac{1}{4\pi\epsilon_0} \frac{q^2}{d} = 9 \times 10^9 \frac{(10 \times 10^{-6})^2}{0.1}\]
03

Calculate the total electrostatic potential energy of the system

As there are 3 pairs of particles, the total electrostatic potential energy of the system (U_total) is obtained by summing the potential energy of each pair: \[U_{total} = 3 \times U_{pair}\] To find the value of \(U_{total}\), we plug the expression for \(U_{pair}\) that we found in step 2: \(U_{total} = 3 \times 9 \times 10^9 \frac{(10 \times 10^{-6})^2}{0.1}\)
04

Calculate the numerical value of the total electrostatic potential energy

Now, we can calculate the numerical value of the total electrostatic potential energy: \(U_{total} = 3 \times 9 \times 10^9 \frac{(10 \times 10^{-6})^2}{0.1} = 27 J\) So, the correct answer is (B) \(27 J\).

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Most popular questions from this chapter

Three charges \(2 q,-q,-q\) are located at the vertices of an equilateral triangle. At the centre of the triangle. (A) The Field is Zero but Potential is non - zero (B) The Field is non - Zero but Potential is zero (C) Both field and Potential are Zero (D) Both field and Potential are non - Zero

Charges of \(+(10 / 3) \times 10^{-9} \mathrm{C}\) are placed at each of the four corners of a square of side \(8 \mathrm{~cm}\). The potential at the intersection of the diagonals is ...... (A) \(150 \sqrt{2}\) Volt (B) \(900 \sqrt{2}\) Volt (C) \(1500 \sqrt{2}\) Volt (D) \(900 \sqrt{2} \cdot \sqrt{2}\) Volt

Four equal charges \(\mathrm{Q}\) are placed at the four corners of a square of each side is ' \(\mathrm{a}\) '. Work done in removing a charge \- Q from its centre to infinity is ....... (A) 0 (B) $\left[\left(\sqrt{2} \mathrm{Q}^{2}\right) /\left(\pi \epsilon_{0} \mathrm{a}\right)\right]$ (C) $\left[\left(\sqrt{2} Q^{2}\right) /\left(4 \pi \epsilon_{0} a\right)\right]$ (D) \(\left[\mathrm{Q}^{2} /\left(2 \pi \epsilon_{0} \mathrm{a}\right)\right]\)

Two positive point charges of \(12 \mu \mathrm{c}\) and \(8 \mu \mathrm{c}\) are placed \(10 \mathrm{~cm}\) apart in air. The work done to bring them $4 \mathrm{~cm}$ closer is (A) Zero (B) \(4.8 \mathrm{~J}\) (C) \(3.5 \mathrm{~J}\) (D) \(-5.8 \mathrm{~J}\)

A charge \(\mathrm{Q}\) is divided into two parts and then they are placed at a fixed distance. The force between the two charges is always maximum when the charges are \(\ldots \ldots\) (A) \((Q / 3),(Q / 3)\) (B) \((\mathrm{Q} / 2),(\mathrm{Q} / 2)\) (C) \((Q / 4),(3 Q / 4)\) (D) \((Q / 5),(4 Q / 5)\)
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