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Chapter 18: Problem 45

ssm Two point charges are located along the \(x\) axis: \(q_{1}=+6.0 \mu \mathrm{C}\) at \(x_{1}=+4.0 \mathrm{cm},\) and \(q_{2}=+6.0 \mu \mathrm{C}\) at \(x_{2}=-4.0 \mathrm{cm} .\) Two other charges are located on the \(y\) axis: \(q_{3}=+3.0 \mu \mathrm{C}\) at \(y_{3}=+5.0 \mathrm{cm},\) and \(q_{4}=-8.0 \mu \mathrm{C}\) at \(y_{4}=+7.0 \mathrm{cm} .\) Find the net electric field (magnitude and direction) at the origin.

Short Answer

Expert verified
The net electric field at the origin is determined by the sum of component vectors, mostly in the y-direction due to symmetry canceling x-components.

Step by step solution

01

Identify Charge Positions

Start by marking the positions of each charge along the coordinate axes. The positive charges \(q_1\) and \(q_2\) are located on the \(x\)-axis at \(x_1 = 4.0\, \text{cm}\) and \(x_2 = -4.0\, \text{cm}\) respectively. The charges \(q_3\) and \(q_4\) are on the \(y\)-axis at \(y_3 = 5.0\, \text{cm}\) and \(y_4 = 7.0\, \text{cm}\).
02

Calculate Electric Field Due to Each Charge

The electric field due to a point charge at a distance \(r\) is given by \(E = \frac{k \cdot |q|}{r^2}\), where \(k = 8.99 \times 10^9\, \text{N m}^2/\text{C}^2\). Calculate \(E_1, E_2, E_3,\) and \(E_4\) for each charge individually. Each field points directly away from each positive charge and towards each negative charge.
03

Determine Directions of Electric Fields

The electric fields due to \(q_1\) and \(q_2\) will both point towards the origin due to symmetry (as both charges are equal and opposite along the \(x\)-axis). The field \(E_3\) due to \(q_3\) will point towards the negative \(y\)-axis, while \(E_4\) due to \(q_4\) will point towards the positive \(y\)-axis due to \(q_4\) being negative.
04

Calculate Net Electric Field Components

Sum up the \(x\)-components and the \(y\)-components of the electric fields separately. Due to symmetry, the \(x\)-components cancel each other out (\(E_{1x} = -E_{2x}\)). Sum the \(y\)-components: \(E_{y} = E_{3y} + E_{4y}\).
05

Find Magnitude and Direction of Net Electric Field

The net electric field \(E_{y}\) becomes the total field, as the \(x\) components cancel. Calculate the magnitude using \(|E| = |E_{y}|\) and because the field is along the \(y\)-axis, the direction is along the \(y\)-axis orientation (based on charge configuration and sign).

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Key Concepts

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

Point Charge
In the realm of electric fields, the concept of a "point charge" is paramount. A point charge is essentially a charge that is concentrated at a single point in space. When we analyze electric fields, we often model charged objects as point charges for simplicity. The rationale is that these charges exhibit a uniform electric field around them.

In the exercise given, each charge is considered to be a point charge located at specific coordinates on the coordinate axes. For instance, the charges given are located precisely at positions such as
  • where \( q_1\) and \( q_2\) are along the \( x\)-axis, and
  • \( q_3\) and \( q_4\) are located on the \( y\)-axis.
By applying this simplification, calculating the electric field at a point (such as the origin) becomes straightforward. Every point charge produces an electric field that weakens with distance according to the inverse square law.
Net Electric Field
The term "net electric field" refers to the total electric field resulting from the combined effect of all individual electric fields present. Each point charge emits its own electric field, and at any given point in space, these fields superpose to create the net electric field.

In our exercise, we are asked to find the net electric field at the origin due to all four point charges. This involves:
  • Calculating the electric field contributed by each charge individually.
  • Determining the effect of these fields along both the x and y axes.
  • Summing these components vectorially to find the cumulative field.
The concept of vector addition is key here, as electric fields have both magnitude and direction. Components along each axis may either add constructively or cancel each other out, as seen when the x-components in the exercise cancel due to symmetry.
Coordinate Axes
When discussing electric fields, utilizing coordinate axes is essential for simplifying calculations and visualizing the problem at hand. The coordinate axes鈥攖ypically the x, y, and z axes鈥攕erve as reference points to pinpoint the locations of charges and the direction of the electric fields.

In the current exercise, the charges are distributed along the x- and y-axes:
  • The charges \(q_1\) and \(q_2\) lie symmetrically along the x-axis.
  • The charges \(q_3\) and \(q_4\) are positioned along the y-axis.
This arrangement helps us decompose each electric field into components that lie neatly along these axes, thereby easing calculations. The goal is to find the net effect of these fields at the origin by summing up component vectors accurately.
Electric Field Direction
Understanding the direction of electric fields produced by point charges is crucial for solving physics problems involving charges. The direction an electric field takes depends on the sign of the charge:
  • A field is directed away from a positive charge.
  • Conversely, it points towards a negative charge.
In this exercise, we consider the vectors for four charges:
  • The fields due to the positive charges \(q_1\) and \(q_2\) are directed towards the origin along the x-axis.
  • For the charge \(q_3\), since it is positive, the field it creates points down along the y-axis.
  • The field associated with the negative charge \(q_4\) moves in the opposite direction to \(q_3\), upwards along the y-axis.
Each electric field's direction influences how the net field is constructed, especially when adding the fields vectorially. Here, because of the relative positions and magnitudes of the charges, the resulting electric field manifests predominantly in the y-direction at the origin.

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Most popular questions from this chapter

An electrically neutral model airplane is flying in a horizontal circle on a 3.0 -m guideline, which is nearly parallel to the ground. The line breaks when the kinetic energy of the plane is 50.0 J. Reconsider the same situation, except that now there is a point charge of \(+q\) on the plane and a point charge of \(-q\) at the other end of the guideline. In this case, the line breaks when the kinetic energy of the plane is \(51.8 \mathrm{J}\). Find the magnitude of the charges.

mmh Two spherical shells have a common center. \(A-1.6 \times 10^{-6}\) C charge is spread uniformly over the inner shell, which has a radius of \(0.050 \mathrm{m} . \mathrm{A}+5.1 \times 10^{-6} \mathrm{C}\) charge is spread uniformly over the outer shell, which has a radius of \(0.15 \mathrm{m}\). Find the magnitude and direction of the electric field at a distance (measured from the common center) of (a) \(0.20 \mathrm{m}\) (b) \(0.10 \mathrm{m},\) and (c) \(0.025 \mathrm{m}\)

An electric field of \(260000 \mathrm{N} / \mathrm{C}\) points due west at a certain spot. What are the magnitude and direction of the force that acts on a charge of \(-7.0 \mu \mathrm{C}\) at this spot?

A circular surface with a radius of \(0.057 \mathrm{m}\) is exposed to a uniform external electric field of magnitude \(1.44 \times 10^{4} \mathrm{N} / \mathrm{C}\). The magnitude of the electric flux through the surface is \(78 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}\). What is the angle (less than \(90^{\circ}\) ) between the direction of the electric field and the normal to the surface?

ssm Two particles, with identical positive charges and a separation of \(2.60 \times 10^{-2} \mathrm{m},\) are released from rest. Immediately after the release, particle 1 has an acceleration \(\overrightarrow{\mathbf{a}}_{1}\) whose magnitude is \(4.60 \times 10^{3} \mathrm{m} / \mathrm{s}^{2},\) while particle 2 has an acceleration \(\overrightarrow{\mathbf{a}}_{2}\) whose magnitude is \(8.50 \times 10^{3} \mathrm{m} / \mathrm{s}^{2}\). Particle 1 has a mass of \(6.00 \times 10^{-6} \mathrm{kg} .\) Find (a) the charge on each particle and (b) the mass of particle 2.
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