Log out Go to App
Go to App

Learning Materials

Features

Discover

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) \(27 J\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two identical metal plates are given positive charges \(\mathrm{Q}_{1}\) and \(\mathrm{Q}_{2}\left(<\mathrm{Q}_{1}\right)\) respectively. If they are now brought close to gather to form a parallel plate capacitor with capacitance \(\mathrm{c}\), the potential difference between them is (A) \(\left[\left(Q_{1}+Q_{2}\right) /(2 c)\right]\) (B) \(\left[\left(\mathrm{Q}_{1}+\mathrm{Q}_{2}\right) / \mathrm{c}\right]\) (C) \(\left[\left(\mathrm{Q}_{1}-\mathrm{Q}_{2}\right) /(2 \mathrm{c})\right]\) (D) \(\left[\left(\mathrm{Q}_{1}-\mathrm{Q}_{2}\right) / \mathrm{c}\right]\)

Two identical charged spheres suspended from a common point by two massless strings of length \(\ell\) are initially a distance d \((d<<\ell)\) apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result the spheres approach each other with a velocity \(\mathrm{v}\). Then function of distance \(\mathrm{x}\) between them becomes \(\ldots \ldots\) (A) \(v \propto x\) (B) \(\mathrm{v} \propto \mathrm{x}^{(-1 / 2)}\) (C) \(\mathrm{v} \propto \mathrm{x}^{-1}\) (D) \(\mathrm{v} \propto \mathrm{x}^{(1 / 2)}\)

Two small conducting sphere of equal radius have charges \(+1 \mathrm{c}\) and \(-2 \mathrm{c}\) respectively and placed at a distance \(\mathrm{d}\) from each other experience force \(F_{1}\). If they are brought in contact and separated to the same distance, they experience force \(F_{2}\). The ratio of \(F_{1}\) to \(F_{2}\) is \(\ldots \ldots \ldots \ldots\) (A) \(-8: 1\) (B) \(1: 2\) (C) \(1: 8\) (D) \(-2: 1\)

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) \(0.85 \times 10^{-6}\)

There are 10 condensers each of capacity \(5 \mu \mathrm{F}\). The ratio between maximum and minimum capacities obtained from these condensers will be (A) \(40: 1\) (B) \(25: 5\) (C) \(60: 3\) (D) \(100: 1\)
See all solutions

Recommended explanations on English Textbooks

Lexis and Semantics Summary

Read Explanation

Syntax

Read Explanation

Language Analysis

Read Explanation

Discourse

Read Explanation

5 Paragraph Essay

Read Explanation

Email

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free