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Chapter 23: Problem 11

Equilateral Triangle Two particles, each with an amount of \(|q|\) equal to charge of \(12 \mathrm{nC}\), are placed at two of the vertices of an equilateral triangle. The length of each side of the triangle is \(2.0 \mathrm{~m}\). What is the magnitude of the electric field at the third vertex of the triangle if (a) both of the charges are positive and (b) one of the charges is positive and the other is negative?

Short Answer

Expert verified
Case (a) The magnitude of the electric field is \(46.7 \text{ N/C}\). Case (b) The magnitude is \(26.97 \text{ N/C}\).

Step by step solution

01

Understand the problem

Two particles with charges \(q = 12 \text{ nC}\) are at two vertices of an equilateral triangle with side length \(2.0 \text{ m}\). Calculate the electric field magnitude at the third vertex for both cases: (a) both charges are positive, and (b) one charge is positive and the other is negative.
02

Identify the formula

The electric field \(E\) due to a point charge is given by \(E = \frac{k q}{r^2}\), where \(k = 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2\) is Coulomb's constant, \(q\) is the charge, and \(r\) is the distance from the charge.
03

Calculate individual electric fields

Each of the charges (\(12 \text{ nC} = 12 \times 10^{-9} \text{ C}\)) creates an electric field at the third vertex. For each charge at distance \(r = 2.0 \text{ m}\), the electric field is \(E_1 = E_2 = \frac{(8.99 \times 10^9) (12 \times 10^{-9})}{(2.0)^2} = 26.97 \text{ N/C}\).
04

Apply superposition principle

The net electric field at the third vertex is the vector sum of the fields due to each charge. In an equilateral triangle, the angle between the electric field vectors due to each charge is \(60^\circ\).
05

Case (a) Both charges are positive

Using vector addition in an equilateral triangle, the resultant field can be found using \(E_{net} = 2E \cos(30^\circ)\), which simplifies to \(E_{net} = 2 \times 26.97 \times \cos(30^\circ) = 46.7 \text{ N/C}\).
06

Case (b) One charge positive, one charge negative

With opposite charges, the fields are in opposite directions. The net electric field \(E_{net} = 2E \sin(30^\circ)\), simplifies to \(E_{net} = 2 \times 26.97 \times \sin(30^\circ) = 26.97 \text{ N/C}\).

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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 helps us understand the interactions between electric charges. According to this principle, the electric force (or field) between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
The equation is:
\[ F = k \frac{{|q_1 q_2|}}{{r^2}} \]
Here,
  1. \( F \) represents the force between the charges.
  2. \( k \) is the Coulomb constant, and its value is approximately \( 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \).
  3. \( q_1 \) and \( q_2 \) are the magnitudes of the charges.
  4. \( r \) is the distance between the charges.

For the electric field \( E \), generated by a single charge at a distance \( r \), we use a similar formula:
\[ E = k \frac{{|q|}}{{r^2}} \]
This provides the magnitude of the electric field created by a point charge.
Superposition principle
The superposition principle is crucial when dealing with multiple charges. It states that the total electric field created by multiple charges is the vector sum of the fields created by each charge separately.
To apply this, we need to:
  • Calculate the electric field due to each individual charge.
  • Consider both the magnitude and the direction of the fields.
  • Add the vector components of these electric fields to find the net electric field.

For example, if you have two charges generating electric fields \( E_1 \) and \( E_2 \), the net electric field at any point is:
\[ \textbf{E}_{\text{net}} = \textbf{E}_1 + \textbf{E}_2 \]
This principle simplifies the complexity of calculations when dealing with multiple sources of electric fields, making complicated systems more manageable.
Electric field vector addition
Adding multiple electric fields is a vital concept in physics. Since electric fields are vectors, we must account for both their magnitudes and directions.
Here's the step-by-step process for vector addition of electric fields:
  • Calculate the electric field created by each charge.
  • Represent each field as a vector, considering its direction.
  • Break down each vector into its components, usually along the x- and y-axes.
  • Add the corresponding components together.
  • Use Pythagorean theorem to find the magnitude of the resultant vector.
  • Determine the direction using trigonometric functions.

For the equilateral triangle example, if the angle between the vectors is \( 60^\text{掳} \), the calculations involve trigonometric functions to combine these fields. For two positive charges, the resultant field is directed midway and simplified as:
\[ E_{\text{net}} = 2E \times \text{cos}(30^\text{掳}) \]
Here, \( E \) is the field from each charge and \(30^\text{掳} \) comes from the triangle's symmetry.
Equilateral triangle geometry
Understanding the geometry of an equilateral triangle is crucial to solving electric field problems involving this shape. An equilateral triangle has all three sides of equal length and all three angles equal to \( 60^\text{掳} \).
In an equilateral triangle, few key aspects are:
  • The distances between each vertex are the same.
  • Height can be found using the Pythagorean theorem.
  • Symmetry can be exploited to simplify electric field calculations.

For our problem:
  • Each side of the triangle is \( 2.0 \text{ m} \).
  • The electric fields from the charges at the vertices will also form an angle of \( 60^\text{掳} \) with each other.
  • The net field direction can be found using vector addition and understanding of symmetry in the electric field orientations.

For the net electric field calculation:
  • If both charges are positive, the fields are combined using cosine of half the angle between them.
  • If one charge is positive and the other negative, fields are combined using sine of half the angle, due to opposite directions.

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Most popular questions from this chapter

Dipole in a Field An electric dipole, consisting of charges of magnitude \(1.50 \mathrm{nC}\) separated by \(6.20 \mu \mathrm{m}\), is in an electric field of strength \(1100 \mathrm{~N} / \mathrm{C}\). (a) What is the magnitude of the electric dipole moment? (b) What is the difference between the potential energies corresponding to dipole orientations parallel to and antiparallel to the field?

Density, Density, Density. (a) A charge of \(-300 e\) is uniformly distributed along a circular arc of radius \(4.00 \mathrm{~cm}\), which subtends an angle of \(40^{\circ} .\) What is the linear charge density along the arc? (b) A charge of \(-300 e\) is uniformly distributed over one face of a circular disk of radius \(2.00 \mathrm{~cm}\). What is the surface charge density over that face? (c) A charge of \(-300 e\) is uniformly distributed over the surface of a sphere of radius \(2.00 \mathrm{~cm} .\) What is the surface charge density over that surface? (d) A charge of \(-300 e\) is uniformly spread through the volume of a sphere of radius \(2.00 \mathrm{~cm} .\) What is the volume charge density in that sphere?

Torque on a Dipole An electric dipole consists of charges \(+2 e\) and \(-2 e\) separated by \(0.78 \mathrm{~nm}\). It is in an electric field of strength \(3.4 \times 10^{6} \mathrm{~N} / \mathrm{C}\). Calculate the magnitude of the torque on the dipole when the dipole moment is (a) parallel to, (b) perpendicular to, and (c) antiparallel to the electric field.

The Size of an Oil Drop In the Millikan oil drop experiment, an atomizer (a sprayer with a fine nozzle) is used to introduce many tiny droplets of oil between two oppositely charged parallel metal plates. Some of the droplets pick up one or more excess electrons. The charge on the plates is adjusted so that the electric force on the excess electrons exactly balances the weight of the droplet. The idea is to look for a droplet that has the smallest electric force and assume that it has only one excess electron. This lets the observer measure the charge on the electron. Suppose we are using an electric field of \(3 \times 10^{4} \mathrm{~N} / \mathrm{C}\). The charge on one electron is about \(1.6 \times\) \(10^{-19} \mathrm{C}\). Estimate the radius of an oil drop whose weight could be balanced by the electric force of this field on one electron.

Dependence of \(E\) Figure \(23-60\) shows a fixed charge (specified by a circle) and a location (specified by the \(\mathrm{x}\) ). \(\mathrm{A}\) test charge is placed at the \(x\) to measure the electric effect of the fixed charge. Complete the following two statements as quantitatively as you can. (For example, if the result is larger by a factor of three, don't say "increases"-say "triples" or "is multiplied by three.") Each statement is meant to be compared with the original situation. (The changes don't accumulate). (a) If the test charge is replaced by one with half the amount of charge, then the electric field it experiences will (b) If the fixed charge is replaced by one with twice the amount of charge, then the electric field experiences by the test charge will
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