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Two \(2.00-\mu \mathrm{C}\) point charges are located on the \(x\) axis. One is at \(x=1.00 \mathrm{m},\) and the other is at \(x=-1.00 \mathrm{m} .\) (a) Determine the electric field on the \(y\) axis at \(y=0.500 \mathrm{m}\) (b) Calculate the electric force on a \(-3.00-\mu \mathrm{C}\) charge placed on the \(y\) axis at \(y=0.500 \mathrm{m}\)

Short Answer

Expert verified
First, calculate the electric fields created by the two charges at the point (0, 0.5) and sum them to get the total electric field at that point. Then, calculate the force on a \(-3.00- \mu C\) charge placed at this point by the calculated electric field.

Step by step solution

01

Calculate Electric Fields

The electric field created by a charge q at a distance r from the charge is given by \( E = k * \frac{|q|}{r^2} \) where k is the Coulomb's constant \((8.99 * 10^9 N m^2/C^2)\). Calculate the magnitudes of electric fields at the point (0, 0.5) created by the charges at (1, 0) and (-1, 0) respectively. Remember to account for directions.
02

Sum the Electric Fields

The total electric field is the vector sum of the two fields. If E1 and E2 are the magnitudes of the electric fields created by the charges at (1, 0) and (-1, 0) respectively, the total electric field at (0, 0.5) is \( E = E1 + E2 \). Since electric field is a vector quantity, take care to sum the components correctly considering the directions.
03

Calculate the Electric Force

The force \(F\) on a charge \(q\) due to an electric field \(E\) is given by \( F = q * E \). Compute the force on a \(-3.00- \mu C\) charge placed at (0, 0.5) by the calculated total electric field. Keep in mind that the force direction will be opposite to the field direction due to the negative charge.

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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 foundational principle in electromagnetism that describes the electric force between two point charges. Formulated by Charles-Augustin de Coulomb in the 18th century, this law states that the force (\( F \)) between two point charges is directly proportional to the product of the magnitudes of the charges (\( q_1 \) and \( q_2 \)) and inversely proportional to the square of the distance (\( r \) that separates them. Mathematically, this can be expressed as: \( F = k \frac{\|q_1q_2\|}{r^2} \) where \( k \) is Coulomb's constant (approximately \( 8.99 \times 10^9 N m^2/C^2 \)).

In the context of the textbook exercise, Coulomb's Law is used to calculate the magnitude of the electric fields that the two point charges on the x-axis generate at a point on the y-axis. Since the charges are equal and positioned symmetrically relative to the y-axis, their fields will be equal in magnitude but will point in opposing directions along the x-axis. Therefore, the electric force they exert on another charge placed on the y-axis can be calculated using Coulomb's Law as well.
Electric Force
The electric force is a vector quantity that represents the interaction between charged objects. According to Coulomb's Law, the electric force exerted by one point charge on another is both attractive and repulsive, depending on whether the charges are of opposite or like signs, respectively.

In our application, after determining the electric fields due to the two charged points using Coulomb's Law, we calculate the electric force on a third charge. The charge, in this case a negative \( -3.00-\mu C \) charge, experiences a force when placed in the electric field created by the other charges. To find this force, we employ the formula \( F = qE \), where \( q \) is the charge on which the force is acting, and \( E \) is the electric field strength at that point. Since the charge here is negative, the direction of the force will be opposite the direction of the electric field, a nuance that's crucial to understanding electric force interactions in electric fields.
Vector Sum of Electric Fields
The vector sum of electric fields is a critical concept that allows us to determine the resultant electric field at a point due to multiple charges. When more than one charge is present, each charge produces its own electric field. To find the total electric field at a given point, we sum up the vector fields produced by all individual charges.

The sum is not a simple arithmetic addition; since we are dealing with vectors, both magnitude and direction must be considered. In our exercise, the charges are placed symmetrically on the x-axis, and their electric fields at a point on the y-axis have equal magnitude but opposite directions. As a result, their horizontal components cancel each other, and we are left with vertical components that add up. By calculating and summing these components, we obtain the total electric field at the point of interest, considering the angles and signs to ensure all vector components are accurately accounted for, leading to a correct calculation of the overall field.

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Most popular questions from this chapter

The electrons in a particle beam each have a kinetic energy \(K\). What are the magnitude and direction of the electric field that will stop these electrons in a distance \(d ?\)

Three solid plastic cylinders all have radius \(2.50 \mathrm{cm}\) and length \(6.00 \mathrm{cm} .\) One (a) carries charge with uniform density \(15.0 \mathrm{nC} / \mathrm{m}^{2}\) everywhere on its surface. Another (b) carries charge with the same uniform density on its curved lateral surface only. The third (c) carries charge with uniform density \(500 \mathrm{nC} / \mathrm{m}^{3}\) throughout the plastic. Find the charge of each cylinder.

The electric field along the axis of a uniformly charged disk of radius \(R\) and total charge \(Q\) was calculated in Example \(23.9 .\) Show that the electric field at distances \(x\) that are large compared with \(R\) approaches that of a point charge \(Q=\sigma \pi R^{2} .\) (Suggestion: First show that \(x /\left(x^{2}+R^{2}\right)^{1 / 2}=\) \(\left(1+R^{2} / x^{2}\right)^{-1 / 2}\) and use the binomial expansion \(\left.(1+\delta)^{n} \approx 1+n \delta \text { when } \delta<<1 .\right)\)

A line of charge starts at \(x=+x_{0}\) and extends to positive infinity. The linear charge density is \(\lambda=\lambda_{0} x_{0} / x\). Determine the electric field at the origin.

(a) Two protons in a molecule are separated by \(3.80 \times\) \(10^{-10} \mathrm{m} .\) Find the electric force exerted by one proton on the other. (b) How does the magnitude of this force compare to the magnitude of the gravitational force between the two protons? (c) What If? What must be the chargeto-mass ratio of a particle if the magnitude of the gravitational force between two of these particles equals the magnitude of electric force between them?

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