Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 1553

Equal charges \(q\) are placed at the vertices \(A\) and \(B\) of an equilateral triangle \(\mathrm{ABC}\) of side \(\mathrm{a}\). The magnitude of electric field at the point \(c\) is \(\ldots \ldots \ldots\) (A) \(\left(\mathrm{Kq} / \mathrm{a}^{2}\right)\) (B) \(\left.(\sqrt{3} \mathrm{Kq}) / \mathrm{a}^{2}\right)\) (C) \(\left.(\sqrt{2} \mathrm{Kq}) / \mathrm{a}^{2}\right)\) (D) $\left[\mathrm{q} /\left(2 \pi \mathrm{t} \varepsilon_{0} \mathrm{a}^{2}\right)\right]$

Short Answer

Expert verified
The magnitude of the electric field at point C is \(\frac{Kq}{a^2}\).

Step by step solution

01

Understanding the principles

The electric field at a point due to a point charge is given by Coulomb's law \(E = K\frac{q}{r^2}\), where \(E\) represents the electric field, \(K\) is Coulomb's constant, \(q\) is the charge and \(r\) is the distance from the charge to the point. It's aimed radially outward for a positive charge and radially inward for a negative charge.
02

Calculate the Electric Field at C due to charge at A

Using Coulombs' law, the electric field at C due to charge q at A is \(E_{A} = \frac{Kq}{a^2}\). This field, due to symmetry, will be perpendicular to AC.
03

Calculate the Electric Field at C due to charge at B

Similarly, the electric field at C due to charge at B is \(E_{B} = \frac{Kq}{a^2}\). This field will also be perpendicular to BC.
04

Total Electric Field at C

The electric fields due to charges at A and B have the same magnitude but are in different directions. Because ABC is an equilateral triangle, the angle ACB (or \(\theta\)) between the two fields is 60 degrees. The resultant or total electric field at C is the vector sum of \(E_A\) and \(E_B\). Using the law of cosines for vector addition, the total electric field at C is given by \[E_C = \sqrt{{E_A}^2 + {E_B}^2 + 2E_AE_B\cos(\pi - \theta)} = \sqrt{{E_A}^2 + {E_B}^2 - 2E_AE_B\cos(\theta)}\]
05

Substitution and simplification

On substituting \(E_A = E_B = E = \frac{Kq}{a^2}\) and \(\cos(\theta) = \cos(60) = \frac{1}{2}\) in the previous step, we get \[E_C = \sqrt{2E^2 - 2E^2\cos(60)} = \sqrt{2E^2 - E^2} = E\sqrt{2 - 1} = E\sqrt{1} = E\] Substitute \(E = \frac{Kq}{a^2}\) back to get the final value as \[E_C = \frac{Kq}{a^2}\] Hence, the magnitude of electric field at point C is \(\frac{Kq}{a^2}\), matching option (A) in the provided answers.

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

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

The capacities of three capacitors are in the ratio \(1: 2: 3\). Their equivalent capacity when connected in parallel is \((60 / 11) \mathrm{F}\) more then that when they are connected in series. The individual capacitors are of capacities in \(\mu \mathrm{F}\) (A) \(4,6,7\) (B) \(1,2,3\) (C) \(1,3,6\) (D) \(2,3,4\)

A charged particle of mass \(\mathrm{m}\) and charge \(q\) is released from rest in a uniform electric field \(E\). Neglecting the effect of gravity, the kinetic energy of the charged particle after 't' second is ...... (A) $\left[\left(\mathrm{Eq}^{2} \mathrm{~m}\right) /\left(2 \mathrm{t}^{2}\right)\right]$ (B) $\left[\left(\mathrm{E}^{2} \mathrm{q}^{2} \mathrm{t}^{2}\right) /(2 \mathrm{~m})\right]$ (C) \(\left[\left(2 \mathrm{E}^{2} \mathrm{t}^{2}\right) /(\mathrm{qm})\right]\) (D) \([(\mathrm{Eqm}) / \mathrm{t}]\)

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 air capacitors \(A=1 \mu F, B=4 \mu F\) are connected in series with $35 \mathrm{~V}\( source. When a medium of dielectric constant \)\mathrm{K}=3$ is introduced between the plates of \(\mathrm{A}\), change on the capacitor changes by (A) \(16 \mu \mathrm{c}\) (B) \(32 \mu \mathrm{c}\) (C) \(28 \mu \mathrm{c}\) (D) \(60 \mu \mathrm{c}\)
See all solutions

Recommended explanations on English Textbooks

Summary Text

Read Explanation

Essay Prompts

Read Explanation

Global English

Read Explanation

Phonetics

Read Explanation

Essay Writing Skills

Read Explanation

Rhetorical Analysis Essay

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