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Chapter 21: Problem 50

A point charge \(q_{1}=-4.00 \mathrm{nC}\) is at the point \(x=0.600 \mathrm{m},\) \(y=0.800 \mathrm{m},\) and a second point charge \(q_{2}=+6.00 \mathrm{nC}\) is at the point \(x=0.600 \mathrm{m}, y=0 .\) Calculate the magnitude and direction of the net electric field at the origin due to these two point charges.

Short Answer

Expert verified
The magnitude is 123.06 N/C, and the direction is 80.1掳 below the negative x-axis.

Step by step solution

01

Understand the Electric Field Due to a Point Charge

The electric field, \(E\), created by a point charge \(q\) at a distance \(r\) is given by Coulomb's law as \(\mathbf{E} = \cfrac{kq}{r^2}\hat{r}\), where \(k = 8.99 \times 10^9 \, \mathrm{N\ m^2/C^2}\) is the Coulomb's constant and \(\hat{r}\) is the unit vector from the charge to the point of interest.
02

Calculate Distance from Charge to Origin

For charge \(q_1\) at \((0.600, 0.800)\) m, the distance \(r_1\) to the origin \((0,0)\) is found using the Pythagorean theorem: \[ r_1 = \sqrt{(0.600^2 + 0.800^2)} = 1.0 \, \text{m} \]For charge \(q_2\) at \((0.600, 0)\) m, the distance \(r_2\): \[ r_2 = 0.600 \, \text{m} \]
03

Calculate Electric Fields at the Origin

The magnitude of the electric field due to \(q_1\), \(E_1\), is:\[ E_1 = \cfrac{k|q_1|}{r_1^2} = \cfrac{8.99 \times 10^9 \times 4.00 \times 10^{-9}}{1.0^2} = 35.96 \, \mathrm{N/C} \]The magnitude of the electric field due to \(q_2\), \(E_2\), is:\[ E_2 = \cfrac{k|q_2|}{r_2^2} = \cfrac{8.99 \times 10^9 \times 6.00 \times 10^{-9}}{0.600^2} = 149.83 \, \mathrm{N/C} \]
04

Determine Directions of Electric Fields

The direction of \(E_1\) is towards \(q_1\), as it is negative, resulting in a vector pointing from the origin to \((0.600, 0.800)\).The direction of \(E_2\) is away from \(q_2\), as it is positive, resulting in a vector pointing from \(0.600, 0\) towards the origin.
05

Resolve Electric Fields into Components

For \(E_1\), resolve into x and y components:\[ E_{1x} = E_1 \cdot \cos(\theta_1) = 35.96 \cdot \cfrac{0.600}{1.0} = 21.58 \, \mathrm{N/C} \]\[ E_{1y} = E_1 \cdot \sin(\theta_1) = 35.96 \cdot \cfrac{0.800}{1.0} = 28.77 \, \mathrm{N/C} \]\(E_1\) points towards the origin, thus the components are negative.\[ E_{1x} = -21.58 \, \mathrm{N/C} \quad E_{1y} = -28.77 \, \mathrm{N/C} \]\(E_2\) acts along the y-axis only since the charge lies on the x-axis directly horizontally aligned with the origin:\[ E_{2x} = 0 \, \mathrm{N/C} \quad E_{2y} = 149.83 \, \mathrm{N/C} \]
06

Calculate Net Electric Field Components

Add the component contributions from \(E_1\) and \(E_2\):\[ E_{net,x} = E_{1x} + E_{2x} = -21.58 + 0 = -21.58 \, \mathrm{N/C} \]\[ E_{net,y} = E_{1y} + E_{2y} = -28.77 + 149.83 = 121.06 \, \mathrm{N/C} \]
07

Compute Magnitude and Direction of Net Electric Field

The magnitude of the net electric field is:\[ E_{net} = \sqrt{E_{net,x}^2 + E_{net,y}^2} = \sqrt{(-21.58)^2 + (121.06)^2} = 123.06 \, \mathrm{N/C} \]The direction \(\theta\) is given by:\[ \theta = \tan^{-1}\left(\cfrac{E_{net,y}}{E_{net,x}}\right) = \tan^{-1}\left(\cfrac{121.06}{-21.58}\right) = -80.1^\circ \]This angle is measured from the positive x-axis, moving clockwise into the fourth quadrant.

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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 fundamental in calculating electric fields due to point charges. This law states that the magnitude of the electric force between two point charges is proportional to the product of the charges and inversely proportional to the square of the distance between them. The formula used is \[ F = \frac{k |q_1 q_2|}{r^2} \]where:- \( F \) is the magnitude of the force between the charges.- \( k \), the Coulomb's constant, is approximately \( 8.99 \times 10^9 \, \mathrm{N\ m^2/C^2} \).- \( |q_1| \) and \( |q_2| \) are the magnitudes of the charges.- \( r \) is the distance between the charges.

In this exercise, this principle helps determine the electric field due to each point charge. We focused on calculating the fields due to each charge at the origin, factoring in distance and the nature, whether attractive or repulsive, of the charges involved.
Point Charge
A point charge is an idealized model of a particle with charge that exists at an infinitely small point in space. It simplifies calculations related to electric fields and forces. Consider the given point charges:- \( q_1 = -4.00 \, \mathrm{nC} \) located at \((0.600, 0.800)\) - \( q_2 = +6.00 \, \mathrm{nC} \) located at \((0.600, 0)\)

The concept of a point charge allows us to apply exact mathematical operations using its positioning in a coordinate system. The position of each charge relative to the origin was crucial to determining the electric field strength and direction because the field strength decreases with the square of the distance, as dictated by coulomb's law.

These charges act as concentrated sources of electric fields, behaving as if their entire charge were located at just one point in space.
Vector Components
Understanding vector components is essential to resolving an electric field into a more manageable form. A vector has both magnitude and direction. Electric fields, being vectors, can be broken down into horizontal (x) and vertical (y) components. This decomposition helps calculate net effects from multiple vectors.
In the exercise:- The electric field from \( q_1 \) was divided into x and y components using the equations: - \( E_{1x} = E_1 \cdot \cos(\theta) \) - \( E_{1y} = E_1 \cdot \sin(\theta) \)

The angle \( \theta \) helps determine the influence of each component. For \( q_2 \), its field acts only in the y-axis, simplifying our calculations. By summing these resolved components, we find the resultant electric field and its direction. Calculating the resultant components from different charges brings clarity to complex systems involving multiple point charges. Properly understanding and resolving vectors ensures accurate field calculations and predictions.

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Most popular questions from this chapter

Three point charges are arranged along the \(x\) -axis. Charge \(q_{1}=+3.00 \mu \mathrm{C}\) is at the origin, and charge \(q_{2}=-5.00 \mu \mathrm{C}\) is at \(x=0.200 \mathrm{m} .\) Charge \(q_{3}=-8.00 \mu \mathrm{C} .\) Where is \(q_{3}\) located if the net force on \(q_{1}\) is 7.00 \(\mathrm{N}\) in the \(-\mathrm{x}\) -direction?

In a rectangular coordinate system a positive point charge \(q=6.00 \times 10^{-9} \mathrm{C}\) is placed at the point \(x=+0.150 \mathrm{m}, y=0,\) and an identical point charge is placed at \(x=-0.150 \mathrm{m}, y=0\) . Find the \(x\) - and \(y\) -components, the magnitude, and the direction of the electric field at the following points: \((a)\) the origin; \((b) x=\) \(0.300 \mathrm{m}, y=0 ;(\mathrm{c}) x=0.150 \mathrm{m}, y=-0.400 \mathrm{m} ;(\mathrm{d}) x=0\) \(y=0.200 \mathrm{m} .\)

A negative point charge \(q_{1}=-4.00 \mathrm{nC}\) is on the \(x\) -axis at \(x=0.60 \mathrm{m} .\) A second point charge \(q_{2}\) is on the \(x\) -axis at \(x=\) \(-1.20 \mathrm{m} .\) What must the sign and magnitude of \(q_{2}\) be for the net electric field at the origin to be (a) 50.0 \(\mathrm{N} / \mathrm{C}\) in the \(+x\) -direction and \((b) 50.0 \mathrm{N} / \mathrm{C}\) in the \(-x\) -direction?

Lightning occurs when there is a flow of electric charge (principally electrons) between the ground and a thundercloud. The maximum rate of charge flow in a lightning bolt is about \(20,000 \mathrm{C} / \mathrm{s} ;\) this lasts for 100\(\mu \mathrm{s}\) or less. How much charge flows between the ground and the cloud in this time? How many electrons flow during this time?

Two very large horizontal sheets are 4.25 \(\mathrm{cm}\) apart and carry equal but opposite uniform surface charge densities of magnitude \(\sigma .\) You want to use these sheets to hold stationary in the region between them an oil droplet of mass 324\(\mu \mathrm{g}\) that carries an excess of five electrons. Assuming that the drop is in vacuum, (a) which way should the electric field between the plates point, and (b) what should \(\sigma\) be?
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