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Chapter 21: Problem 88

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?

Short Answer

Expert verified
(a) +2.40 nC (b) -0.80 nC

Step by step solution

01

Understanding the Setup

We have two point charges, \( q_1 = -4.00 \text{ nC} \) at \( x = 0.60 \text{ m} \) and \( q_2 \) at \( x = -1.20 \text{ m} \). The goal is to calculate the electric field at the origin produced by these charges, which currently are unknown for \( q_2 \)'s sign and magnitude. The electric field due to a point charge is given by the formula: \( E = \frac{k |q|}{r^2} \), where \( k \) is Coulomb's constant \( 8.99 \times 10^9 \text{ N路m}^2/\text{C}^2 \) and \( r \) is the distance from the charge to the field point.
02

Calculating Electric Field at the Origin by \( q_1 \)

At the origin, the electric field due to \( q_1 \) will point towards \( q_1 \) along the \( x \)-axis because \( q_1 \) is negative. The distance from \( q_1 \) to the origin is \( 0.60 \text{ m} \). Thus, the magnitude of the electric field due to \( q_1 \) at the origin is given by:\[E_1 = \frac{k |q_1|}{(0.60)^2}\]Substituting the values gives:\[E_1 = \frac{8.99 \times 10^9 \times 4.00 \times 10^{-9}}{0.36} = 99.89 \text{ N/C} \] This electric field is directed towards the negative \( x \)-direction.
03

Setting Up Equations for Net Electric Field

For part (a), the resultant electric field needs to be \( 50.0 \text{ N/C} \) towards the \( +x \)-direction, meaning \( E_2 - E_1 = 50.0 \). For part (b), the resultant needs to be \( 50.0 \text{ N/C} \) towards the \( -x \)-direction, which gives the equation \( E_1 - E_2 = 50.0 \) N/C. Let's solve them accordingly.
04

Solving for \( q_2 \) in Part (a)

For part (a), we need to solve:\[E_2 - 99.89 = 50.0\]So, \( E_2 = 149.89 \text{ N/C} \). Since the distance from \( q_2 \) to the origin is \( 1.20 \text{ m} \), we use:\[E_2 = \frac{k |q_2|}{(1.20)^2} \Rightarrow |q_2| = \frac{E_2 \times (1.20)^2}{k}\]Substituting the values gives:\[|q_2| = \frac{149.89 \times 1.44}{8.99 \times 10^9} = 2.40 \text{ nC}\]The direction of \( E_2 \) towards the origin implies \( q_2 \) must be positive.
05

Solving for \( q_2 \) in Part (b)

For part (b), we need to solve:\[99.89 - E_2 = 50.0\]So, \( E_2 = 49.89 \text{ N/C} \). Using the same electric field formula:\[E_2 = \frac{k |q_2|}{(1.20)^2} \Rightarrow |q_2| = \frac{E_2 \times (1.20)^2}{k}\]Substituting the values gives:\[|q_2| = \frac{49.89 \times 1.44}{8.99 \times 10^9} = 0.80 \text{ nC}\]The direction of \( E_2 \) away from the origin implies \( q_2 \) must be negative.

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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 understanding electric fields around point charges. It tells us how two charges interact with each other. The law states that the electric force \( F \) between two point charges \( q_1 \) and \( q_2 \) is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance \( r \) between them. This relationship is given by: \[ F = k \frac{|q_1 q_2|}{r^2} \] where \( k \) is Coulomb's constant, \( 8.99 \times 10^9 \text{ N路m}^2/\text{C}^2 \). - Both charges can be positive, negative, or mixed: - If both charges are of the same sign, they repel each other. - If they have opposite signs, they attract each other.- The force \( F \) is a vector, meaning it has both magnitude and direction.Using this law, we can determine the direction of the electric field a charge creates, which is essential for predicting how charges will move when placed close together.
Point Charges
A point charge is a charged object that creates an electric field around it, affecting other charges within its vicinity. Point charges are often considered in theoretical studies because they simplify calculations of electric fields and forces. Point charges may be negative or positive, determining the direction of the electric field they produce.- For negative point charges, the electric field is directed towards the charge.- For positive point charges, the electric field is directed away from the charge.An example to illustrate point charges would be \( q_1 = -4.00 \text{ nC} \) in the given exercise. It exerts an electric influence on surrounding areas. When solving problems involving point charges, knowing the sign of the charge helps determine how they influence other charges or electric fields around them.
Electric Field Direction
The direction of the electric field created by a charge is a fundamental idea when evaluating the net electric field at a particular point. The electric field direction is determined by the nature of the charge:- For a negative charge, the electric field lines point towards the charge.- For a positive charge, these lines point away from the charge.In our exercise, the electric field due to \( q_1 \) at the origin points towards \( q_1 \) because of its negative sign. If we introduce another charge \( q_2 \) , determining the combined effect involves evaluating contributions from each charge. - If the net electric field needs to point in the \(+x\) direction, the contribution from \( q_2 \) must outbalance the effect of \( q_1 \).- Conversely, if the field is to point in the \(-x\) direction, \( q_2 \)'s contribution must be lesser in magnitude. Understanding electric field direction helps in designing settings where specific electric field configurations are necessary, such as in cathode ray tubes or particle accelerators.

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Most popular questions from this chapter

A charge \(q_{1}=+5.00 \mathrm{nC}\) is placed at the origin of an \(x y\) - coordinate system, and a charge \(q_{2}=-2.00 \mathrm{nC}\) is placed on the positive \(x\) -axis at \(x=4.00 \mathrm{cm} .\) (a) If a third charge \(q_{3}=+6.00 \mathrm{nC}\) is now placed at the point \(x=4.00 \mathrm{cm}, y=3.00 \mathrm{cm},\) find the \(x\) - and \(y\) -components of the total force exerted on this charge by the other two. (b) Find the magnitude and direction of this force.

Point charges \(q_{1}=-4.5 \mathrm{nC}\) and \(q_{2}=+4.5 \mathrm{nC}\) are separated by \(3.1 \mathrm{mm},\) forming an electric dipole. (a) Find the electric dipole moment (magnitude and dircction). (b) The charges are in a uniform electric field whose direction makes an angle of \(36.9^{\circ}\) with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude \(7.2 \times 10^{-9} \mathrm{N} \cdot \mathrm{m} ?\)

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?

(a) Assuming that only gravity is acting on it, how far does an electron have to be from a proton so that its acceleration is the same as that of a freely falling object at the earth's surface? (b) Suppose the earth were made only of protons but had the same size and mass it presently has. What would be the acceleration of an electron released at the surface? Is it necessary to consider the gravitational attraction as well as the electrical force? Why or why not?

A particle has charge \(-3.00 \mathrm{nC}\) (a) Find the magnitude and direction of the electric field due to this particle at a point 0.250 \(\mathrm{m}\) directly above it. (b) At what distance from this particle does its electric field have a magnitude of 12.0 \(\mathrm{N} / \mathrm{C} ?\)
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