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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.

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Most popular questions from this chapter

The electric Potential \(\mathrm{V}\) at any Point $0(\mathrm{x}, \mathrm{y}, \mathrm{z}\( all in meters \))\( in space is given by \)\mathrm{V}=4 \mathrm{x}^{2}\( volt. The electric field at the point \)(1 \mathrm{~m}, 0.2 \mathrm{~m})\( in volt meter is \)\ldots \ldots .$ (A) 8 , along negative \(\mathrm{x}\) - axis (B) 8 , along positives \(\mathrm{x}\) - axis (C) 16 , along negative \(\mathrm{x}\) -axis (D) 16 , along positives \(\mathrm{x}\) -axis

A ball of mass \(1 \mathrm{gm}\) and charge \(10^{-8} \mathrm{c}\) moves from a point \(\mathrm{A}\), where the potential is 600 volt to the point \(B\) where the potential is zero. Velocity of the ball of the point \(\mathrm{B}\) is $20 \mathrm{~cm} / \mathrm{s}\(. The velocity of the ball at the point \)\mathrm{A}$ will be \(\ldots \ldots\) (A) \(16.8(\mathrm{~m} / \mathrm{s})\) (B) \(22.8(\mathrm{~cm} / \mathrm{s})\) (C) \(228(\mathrm{~cm} / \mathrm{s})\) (D) \(168(\mathrm{~m} / \mathrm{s})\)

A simple pendulum of period \(\mathrm{T}\) has a metal bob which is negatively charged. If it is allowed to oscillate above a positively charged metal plate, its period will ...... (A) Remains equal to \(\mathrm{T}\) (B) Less than \(\mathrm{T}\) (C) Infinite (D) Greater than \(\mathrm{T}\)

Large number of capacitors of rating $10 \mu \mathrm{F} / 200 \mathrm{~V} \mathrm{~V}$ are available. The minimum number of capacitors required to design a \(10 \mu \mathrm{F} / 700 \mathrm{~V}\) capacitor is (A) 16 (B) 8 (C) 4 (D) 7

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) \(-5.8 \mathrm{~J}\)
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