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Three particles, each with positive charge \(Q\), form an equilateral triangle, with each side of length \(d\). What is the magnitude of the electric field produced by the particles at the midpoint of any side?

Short Answer

Expert verified
The electric field magnitude at the midpoint is \(\frac{16kQ}{3d^2}\).

Step by step solution

01

Analyze the Triangle

Consider the equilateral triangle with charges at the vertices A, B, and C. Each side of the triangle has length \(d\). We are interested in finding the electric field at the midpoint of one of these sides, say AB.
02

Calculate Individual Electric Field Contributions

The electric field produced by a charge \(Q\) at a distance \(r\) is \(E = \frac{kQ}{r^2}\), where \(k\) is Coulomb's constant. First, calculate the electric field at the midpoint M due to A and B, both charges being \(Q\) and M at distance \(\frac{d}{2}\) from each charge: \(E_A = E_B = \frac{kQ}{(d/2)^2} = \frac{4kQ}{d^2}\).
03

Resolve Electric Field Components

Interpret the symmetry. The fields \(E_A\) and \(E_B\) are equal in magnitude and have equal angles with the perpendicular line originating from the midpoint M and going into the plane of the triangle. Thus, their horizontal components cancel each other, and we only need the vertical components.
04

Calculate Net Electric Field Components

Decompose \(E_A\) and \(E_B\) each into vertical components using \(\cos(30^\circ)\) since the midpoint divides the triangle into two right triangles. The component perpendicular to AB is: \(E_{A_y} = E_{B_y} = \frac{4kQ}{d^2} \cdot \cos(30^\circ) = \frac{4kQ}{d^2} \cdot \frac{\sqrt{3}}{2}\). The net field from A and B is then: \(E_{net} = 2 \times \frac{4kQ}{d^2} \cdot \frac{\sqrt{3}}{2} = \frac{4\sqrt{3}kQ}{d^2}\).
05

Include the Contribution from the Third Charge

The midpoint M is directly perpendicular to side AB and at a distance \(\frac{\sqrt{3}d}{2}\) from the third charge C (using 30-60-90 triangle relationships). The field due to C is: \(E_C = \frac{kQ}{\left(\frac{\sqrt{3}d}{2}\right)^2} = \frac{4kQ}{3d^2}\).
06

Combine Electric Field Magnitudes

The net electric field at M is the vector addition of the vertical component from A and B and the field from C, which both point in the upward perpendicular direction. Thus the total magnitude is: \(E = \frac{4\sqrt{3}kQ}{d^2} + \frac{4kQ}{3d^2}\).
07

Simplify the Expression

Calculate the total magnitude: \(E = \frac{12kQ}{3d^2} + \frac{4kQ}{3d^2} = \frac{16kQ}{3d^2}\).

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Key Concepts

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

Coulomb's Law
Coulomb's Law is a fundamental principle in physics that describes the force between two charges. It tells us that the electric force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them. This is expressed as \[ F = \frac{k \cdot Q_1 \cdot Q_2}{r^2} \]where:
  • \( k \) is Coulomb's constant (approximately \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \))
  • \( Q_1 \) and \( Q_2 \) are the magnitudes of the charges
  • \( r \) is the distance between the charges
Coulomb's Law helps us determine not only the magnitude of the electric force but also the electric field produced by a single point charge. The electric field \( E \), at a distance \( r \) from a charge \( Q \), is given by \[ E = \frac{kQ}{r^2} \] This principle is crucial when dealing with multiple charges, such as in the configuration of charges at the vertices of an equilateral triangle. By applying Coulomb's Law, we can calculate how these charges interact with each other.
Equilateral Triangle
An equilateral triangle has three sides of equal length and three angles each measuring \( 60^\circ \). In problems involving electric fields and charges, this symmetry often plays a crucial role. For example, with three equal charges placed at each vertex of the triangle, we can intuitively expect symmetry to affect how the electric fields interact.

In the exercise we're examining, the goal is to find the electric field at the midpoint of a side of this triangle. The symmetric nature simplifies calculations because the contributions of equal charges at equal distances are equal in magnitude. This symmetry allows for the cancellation of certain components, making the net electric field easier to compute. Recognizing and using symmetry greatly reduces the complexity of the problem.
Charge Distribution
A charge distribution refers to how electrical charges are arranged in a system, which significantly impacts the resultant electric field. In the case of an equilateral triangle with three identical charges \(Q\) at each vertex, the distribution is uniform.

This uniformity means that at any given side midpoint, the influences of two adjacent charges are identical, leading to cancellation or reinforcement of certain field components. The uniform distribution also dictates that any non-perpendicular distance, such as from the third charge, needs special geometric considerations when calculating its contribution.
  • This ensures that calculations involving uniform charges remain consistent among multiple triangle configurations.
  • Uniform distributions simplify the mathematical expressions, as the symmetry can be leveraged to find a solution without needing extensive calculation for each individual charge.
Electric Field Components
The electric field, a vector quantity, is made up of components that can be analyzed separately. In the context of calculating fields in geometric shapes like triangles, understanding components is essential. Each charge contributes to the total electric field at a point with both direction and magnitude.

For an equilateral triangle, if considering the electric field at the midpoint of a side:
  • The vertical components from the adjacent charges reinforce each other due to symmetry.
  • The horizontal components cancel each other out, simplifying the net calculation.
  • The contribution from the opposite vertex charge must be calculated distinctly, often considering a different geometric distance.
Breaking down each charge's electric field into perpendicular components allows us to focus on how they add up. The net electric field is then the vector sum of these components, which often only leaves a significant contribution in one direction, reflecting the symmetry involved in these problems.

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Most popular questions from this chapter

Two charged particles are attached to an \(x\) axis: Particle 1 of charge \(-2.00 \times 10^{-7} \mathrm{C}\) is at position \(x=6.00 \mathrm{~cm}\) and particle 2 of charge \(+2.00 \times 10^{-7} \mathrm{C}\) is at position \(x=21.0 \mathrm{~cm} .\) Midway between the particles, what is their net electric field in unit-vector notation?

How much work is required to turn an electric dipole \(180^{\circ}\) in a uniform electric field of magnitude \(E=46.0 \mathrm{~N} / \mathrm{C}\) if the dipole moment has a magnitude of \(p=3.02 \times\) \(10^{-25} \mathrm{C} \cdot \mathrm{m}\) and the initial angle is \(64^{\circ} ?\)

The following table gives the charge seen by Millikan at different times on a single drop in his experiment. From the data, calculate the elementary charge \(e\). $$ \begin{array}{lll} \hline 6.563 \times 10^{-19} \mathrm{C} & 13.13 \times 10^{-19} \mathrm{C} & 19.71 \times 10^{-19} \mathrm{C} \\ 8.204 \times 10^{-19} \mathrm{C} & 16.48 \times 10^{-19} \mathrm{C} & 22.89 \times 10^{-19} \mathrm{C} \\ 11.50 \times 10^{-19} \mathrm{C} & 18.08 \times 10^{-19} \mathrm{C} & 26.13 \times 10^{-19} \mathrm{C} \\ \hline \end{array} $$

A particle of charge \(-q_{1}\) is at the origin of an \(x\) axis (a) At what location on the axis should a particle of charge \(-4 q_{1}\) be placed so that the net electric field is zero at \(x=2.0 \mathrm{~mm}\) on the axis? (b) If, instead, a particle of charge \(+4 q_{1}\) is placed at that location, what is the direction (relative to the positive direction of the \(x\) axis) of the net electric field at \(x=2.0 \mathrm{~mm} ?\)

A \(10.0 \mathrm{~g}\) block with a charge of \(+8.00 \times 10^{-5} \mathrm{C}\) is placed in an electric field \(\vec{E}=(3000 \hat{\mathrm{i}}-600 \mathrm{j}) \mathrm{N} / \mathrm{C}\). What are the (a) magnitude and (b) direction (relative to the positive direction of the \(x\) axis) of the electrostatic force on the block? If the block is released from rest at the origin at time \(t=0\), what are its (c) \(x\) and (d) \(y\) coordinates at \(t=3.00 \mathrm{~s}\) ?

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