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Equal charges \(q\) each are placed at the vertices \(A\) and \(B\) of an equilateral triangle \(A B C\) of side \(a\). The magnitude of electric field intensity at the point \(C\) is (a) \(\frac{q}{4 \pi \varepsilon_{0} a^{2}}\) (b) \(\frac{\sqrt{2 q}}{4 \pi \varepsilon_{0} a^{2}}\) (c) \(\frac{q \sqrt{3}}{4 \pi \varepsilon_{0} a^{2}}\) (d) \(\frac{2 q}{4 \pi \varepsilon_{0} a^{2}}\)

Short Answer

Expert verified
The correct answer is (c) \(\frac{q \sqrt{3}}{4 \pi \varepsilon_0 a^2}\).

Step by step solution

01

Understand the Electric Field at a Point

The electric field \(E\) due to a point charge \(q\) at a distance \(r\) is given by the formula \[E = \frac{q}{4 \pi \varepsilon_0 r^2}\]. Here, \(\varepsilon_0\) is the permittivity of free space. Since the points are on the vertices of an equilateral triangle, the distances \(AC\) and \(BC\) are equal to the side length \(a\).
02

Calculate the Electric Field Due to Charge at A

The electric field at point \(C\) due to the charge at point \(A\) is directed along \(CA\). Using the formula for electric field, \[E_A = \frac{q}{4 \pi \varepsilon_0 a^2}\].
03

Calculate the Electric Field Due to Charge at B

The electric field at point \(C\) due to the charge at point \(B\) is directed along \(CB\). Similarly, \[E_B = \frac{q}{4 \pi \varepsilon_0 a^2}\].
04

Find the Net Effective Electric Field at Point C

Since both electric fields \(E_A\) and \(E_B\) have the same magnitude \(\left( \frac{q}{4 \pi \varepsilon_0 a^2} \right)\) but are directed at a 60-degree angle to each other due to the equilateral triangle arrangement, use vector addition. The resultant electric field can be found using the formula: \[E_{net} = \sqrt{E_A^2 + E_B^2 + 2 E_A E_B \cos(60^\circ)} = \sqrt{2 \left(\frac{q}{4 \pi \varepsilon_0 a^2}\right)^2 + \left(\frac{q}{4 \pi \varepsilon_0 a^2}\right)^2}\].
05

Simplify and Compute the Resultant Field Intensity

Simplifying the expression, calculate the effective electric field: \[E_{net} = \sqrt{2 \left(\frac{q^2}{16 \pi^2 \varepsilon_0^2 a^4}\right) + \frac{q^2}{16 \pi^2 \varepsilon_0^2 a^4}} = \sqrt{3} \left(\frac{q}{4 \pi \varepsilon_0 a^2}\right)\]. So, the correct answer is option (c) \(\frac{q \sqrt{3}}{4 \pi \varepsilon_0 a^2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilateral Triangle
An equilateral triangle is a triangle in which all three sides are equal in length, and all three angles are equal, each measuring 60 degrees. This symmetry plays a significant role in problems involving electric fields, especially when charges are placed at the vertices of the triangle. In this specific exercise, charges of equal magnitude are positioned at two corners of the triangle, creating a situation where their electric field effects at the third point (the corner where there is no charge) need to be combined.

Understanding the properties of an equilateral triangle helps in visualizing the problem. The distances from point C to points A and B are equal, simplifying the calculations for each electric field component from the charged vertices. Knowing that the angles are 60 degrees allows us to correctly apply vector addition to find the resultant electric field at point C.
Electric Field Formula
The electric field intensity due to a point charge is a measure of the force a charge would experience if placed in the vicinity of another charge. It is calculated using the formula:
  • \[E = \frac{q}{4 \pi \varepsilon_0 r^2}\]
This formula tells you that the field intensity is directly proportional to the magnitude of the charge and inversely proportional to the square of the distance from the charge. The constant \(\varepsilon_0\) is known as the permittivity of free space, a fundamental physical constant that mediates the electric force in a vacuum.

In our exercise, this formula is used twice, for the charges at points A and B, to calculate the electric fields impacting point C. Since C is equally distant from both A and B (because the triangle is equilateral), the field intensities calculated separately for each charge are the same in magnitude. Knowing how to apply this formula is crucial for solving the problem effectively.
Vector Addition
Vector addition is a critical technique for finding the resultant of multiple vectors acting on a single point. Each vector represents a quantity having both magnitude and direction. When calculating the net electric field at point C, we recognize that both electric field vectors \(E_A\) and \(E_B\) are equal in magnitude but directed at an angle to each other.

The operation used to combine these vectors can be described by the following formula:
  • \[E_{net} = \sqrt{E_A^2 + E_B^2 + 2 E_A E_B \cos(60^\circ)}\]
Notice the term \(\cos(60^\circ)\), which arises due to the 60-degree angle between the vectors. This formula is derived from the law of cosines, applicable in vector addition whenever vectors do not align along the same line.Understanding vector addition allows for the calculation of the effective electric field, or \(E_{net}\), acting at point C. The symmetry of the equilateral triangle simplifies this task, but recognizing and applying vector addition correctly is vital to arrive at the solution, which is ultimately simplified to \(\frac{q \sqrt{3}}{4 \pi \varepsilon_0 a^2}\).

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Most popular questions from this chapter

Two charges \(5 \times 10^{-8} \mathrm{C}\) and \(-3 \times 10^{-8} \mathrm{C}\) are located \(16 \mathrm{~cm}\) apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero. [NCERT] (a) \(6 \mathrm{~cm}\) from the charge \(-3 \times 10^{-8} \mathrm{C}\) (b) \(6 \mathrm{~cm}\) from the charge \(5 \times 10^{-8} \mathrm{C}\) (c) \(9 \mathrm{~cm}\) from the charge \(-3 \times 10^{-8} \mathrm{C}\) (d) \(9 \mathrm{~cm}\) from the charge \(5 \times 10^{-8} \mathrm{C}\)

A positively charged particle is released from rest in an uniform electric field. The electric potential energy of the charge [NCERT Exemplar] (a) remains a constant because the electric field is uniform (b) increases because the charge moves along the electric field (c) decreases because the charge moves along the electric field (d) decreases because the charge moves opposite to the electric field

Work done in carrying a charge \(Q_{1}\) once round a circle of radius \(R\) with a charge \(Q_{2}\) at the centre is (a) \(\frac{Q_{1} Q_{2}}{4 \pi \varepsilon_{0} R^{2}}\) (b) zero (c) \(\frac{Q_{1} Q_{2}}{4 \pi \varepsilon_{0} R}\) (d) infinite

A solid conducting sphere having a charge \(Q\) is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be \(V\). If the shell is now given a charge \(-3 Q\), the new potential difference between the same two surfaces is (a) \(\underline{V}\) (b) \(2 \mathrm{~V}\) (c) \(4 V\) (d) \(-2 V\)

A \(600 \mathrm{pF}\) capacitor is charged by a \(200 \mathrm{~V}\) supply. Then, it is disconnected from the supply and is connected to another uncharged \(600 \mathrm{pF}\) capacitor. How much electrostatic energy is lost in the process? [NCERT] (a) \(4 \times 10^{-6} \mathrm{~J}\) (b) \(6 \times 10^{-6} \mathrm{~J}\) (c) \(5 \times 10^{-6} \mathrm{~J}\) (d) \(8 \times 10^{-6} \mathrm{~J}\)

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