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

Short Answer

Expert verified
(a) E_net = E_1; (b) E_net = 0.

Step by step solution

01

Understand the Problem

We need to calculate the electric field at the third vertex of an equilateral triangle where two charged particles are placed at the other vertices. The charges are both 12 nC, and the distance between every pair of vertices is 2.0 m.
02

Setup the Electric Field Formula

The electric field due to a single point charge is given by the formula: \[E = \frac{k \cdot |Q|}{r^2}\]where \(E\) is the electric field, \(k\) is Coulomb's constant \( (8.99 \times 10^9 \text{ N m}^2/ ext{C}^2) \), \(Q\) is the charge, and \(r\) is the distance from the charge.
03

Calculate the Electric Field from Each Charge (Part a)

For part (a), both charges are positive. Calculate the electric field at the third vertex due to each charge using the formula from Step 2:\[ E_1 = E_2 = \frac{8.99 \times 10^9 \text{ N m}^2/ ext{C}^2 \times 12 \times 10^{-9} \text{ C}}{2.0^2 \text{ m}^2} \]Solve for \(E_1\) and \(E_2\). Both fields have the same magnitude and form an angle of 60° with respect to each other due to the equilateral triangle.
04

Calculate Net Electric Field (Part a)

To find the net electric field when both charges are positive, break the two electric field vectors into their x and y components and sum them:\[ E_{net,x} = E_1 \cos(60^{\circ}) + E_2 \cos(60^{\circ}) \]\[ E_{net,y} = E_1 \sin(60^{\circ}) - E_2 \sin(60^{\circ}) \]Because the y-components are equal and opposite, they cancel each other out, leading to:\[ E_{net} = 2 \cdot E_1 \cdot \cos(60^{\circ}) = E_1 \]Calculate the magnitude of \(E_{net}\).
05

Analyze the Configuration (Part b)

In part (b), one charge is positive and the other is negative. The directions of the electric fields at the third vertex due to each charge will now be along the line joining the vertex from each charge directly.
06

Calculate Net Electric Field (Part b)

Since the configurations point in the same line but in opposite directions (along the line of joining), to find the net electric field at the vertex we simply subtract their magnitudes:\[ E_{net} = E_1 + (-E_2) = E_1 - E_2 \]Since \(E_1 = E_2\), the net electric field is:\[ E_{net} = 0 \]Thus the net electric field at the third vertex is zero in this scenario.

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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 that describes the force between two point charges. It states that the electric force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. This law can be mathematically expressed as:
  • Formula: \( F = \frac{k \cdot |Q_1| \cdot |Q_2|}{r^2} \)
  • Where \( F \) is the force between the charges, \( Q_1 \) and \( Q_2 \) are the magnitudes of the charges, \( r \) is the distance between the charges, and \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \).
Coulomb's Law not only applies to the forces between the charges but is also instrumental in calculating the electric field generated by a point charge. The electric field due to a point charge can be found using a similar formulation:
  • Formula: \( E = \frac{k \cdot |Q|}{r^2} \)
  • Where \( E \) is the electric field, \( Q \) is the charge, and \( r \) is the distance from the charge.
This principle is pivotal in solving the exercise as it allows us to determine the influence each charge exerts on a point in space—namely, the third vertex of the equilateral triangle. The electric field from each point charge acts as a vector, which means its direction and magnitude must be considered to find the resultant field when multiple charges are involved.
Equilateral Triangle
An equilateral triangle is a triangle in which all three sides are of equal length, and all three internal angles are equal, each being 60°. These properties make equilateral triangles quite symmetrical, which simplifies calculations in physics problems involving geometry, like the one in this exercise.In the context of electric fields, the symmetry of an equilateral triangle makes it easier to break down vector components and perform vector addition. For example, when calculating the net electric field at a vertex of the triangle where charges are placed at the other two vertices, the symmetry ensures that the electric fields from the charges are of the same magnitude, provided the charges are identical.The angles of 60° play a critical role when resolving vectors into components. Because these angles are equal, it's easy to predict the behavior of the vector components:
  • The x-component of an electric field vector at the third vertex is found by multiplying the field magnitude by \( \cos(60^{\circ}) \).
  • The y-component is found by multiplying the field magnitude by \( \sin(60^{\circ}) \).
This consistency allows physicists and students alike to rely on geometric symmetry to simplify otherwise complex vector mathematics.
Point Charges
Point charges are a theoretical concept used to represent charged objects that are small enough so that their size does not significantly affect the electric field or the force calculations. In essence, we consider the whole charge to be concentrated at a single point in space.The exercise involving two point charges at vertices of an equilateral triangle calls upon this concept. By treating the charged particles as point charges, we can directly apply formulas and principles such as Coulomb's Law to find the electric field at any given point away from the charges.
  • Features of point charges include:
    • Defined by their charge magnitude (e.g., 12 nC in the exercise).
    • Their influence (electric field) diminishes with distance, specifically following an inverse square law calculated by \( E = \frac{k \cdot |Q|}{r^2} \).
  • Applications:
    • Modeling atoms in a molecule where electrons and protons are treated as point charges.
    • Understanding the distribution of electric charge in conductors and insulators.
In the exercise scenario, viewing the particles as point charges allows us to deduce the electric field at the third vertex by calculating the contributions from each charge independently, and then using vector addition to find the net field influenced by both charges.

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Most popular questions from this chapter

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 \(e\). $$ \begin{array}{lll} \hline 6.563 \times 10^{-19} \mathrm{C} & 13.13 \times 10^{-19} \mathrm{C} & 19.71 \times 10^{-19} \mathrm{C} \\ 8.204 \times 10^{-19} \mathrm{C} & 16.48 \times 10^{-19} \mathrm{C} & 22.89 \times 10^{-19} \mathrm{C} \\ 11.50 \times 10^{-19} \mathrm{C} & 18.08 \times 10^{-19} \mathrm{C} & 26.13 \times 10^{-19} \mathrm{C} \\ \hline \end{array} $$

Humid air breaks down (its molecules become ionized) in an electric field of \(3.0 \times 10^{6} \mathrm{~N} / \mathrm{C}\). In that field, what is the magnitude of the electrostatic force on (a) an electron and (b) an ion with a single electron missing?

An electron on the axis of an electric dipole is \(25 \mathrm{~nm}\) from the center of the dipole. What is the magnitude of the electrostatic force on the electron if the dipole moment is \(3.6 \times 10^{-29} \mathrm{C} \cdot \mathrm{m} ?\) Assume that \(25 \mathrm{~nm}\) is much larger than the separation of the charged particles that form the dipole.

An electric dipole with dipole moment $$ \vec{p}=(3.00 \hat{\mathrm{i}}+4.00 \hat{\mathrm{j}})\left(1.24 \times 10^{-30} \mathrm{C} \cdot \mathrm{m}\right) $$ is in an electric field \(\vec{E}=(4000 \mathrm{~N} / \mathrm{C}) \hat{\mathrm{i}}\). (a) What is the potential energy of the electric dipole? (b) What is the torque acting on it? (c) If an external agent turns the dipole until its electric dipole moment is $$ \vec{p}=(-4.00 \hat{\mathrm{i}}+3.00 \hat{\mathrm{j}})\left(1.24 \times 10^{-30} \mathrm{C} \cdot \mathrm{m}\right) $$ how much work is done by the agent?

At some instant the velocity components of an electron moving between two charged parallel plates are \(v_{x}=1.5 \times 10^{5} \mathrm{~m} / \mathrm{s}\) and \(v_{y}=3.0 \times 10^{3} \mathrm{~m} / \mathrm{s}\). Suppose the electric field between the plates is uniform and given by \(\vec{E}=(120 \mathrm{~N} / \mathrm{C}) \hat{j} .\) In unit-vector notation, what are (a) the electron's acceleration in that field and (b) the electron's velocity when its \(x\) coordinate has changed by \(2.0 \mathrm{~cm}\) ?

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