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Chapter 1: Problem 9

Two point charges are located on the \(y\)-axis as follows: charge \(q_{1}=-1.50 \mathrm{nC}\) at \(y=-0.600 \mathrm{~m}\), and charge \(q_{2}=\) \(+3.20 \mathrm{nC}\) at the origin \((y=0)\). What is the total force (magnitude and direction) exerted by these two charges on a third charge \(q_{3}=+5.00 \mathrm{nC}\) located at \(y=-0.400 \mathrm{~m} ?\)

Short Answer

Expert verified
The net force on \(q_3\) is \(7.91 \times 10^{-7} \, N\) downwards along the y-axis.

Step by step solution

01

Understand the Problem

We need to calculate the force exerted by two charges on a third charge using Coulomb's Law. The two given charges are on the y-axis at specific positions, and we want to determine the net force on the third charge, both in magnitude and direction.
02

Recall Coulomb's Law

Coulomb's Law states that the force between two point charges is given by: \[ F = k \frac{|q_1 q_2|}{r^2} \]where \(k = 8.99 \times 10^9 \mathrm{~N~m^2/C^2}\) is the electrostatic constant, \(q_1\) and \(q_2\) are the charges, and \(r\) is the distance between the charges.
03

Calculate Distance Between Charges

Find the distances between the third charge \(q_3\) located at \(y = -0.400 \mathrm{~m}\) and each of the other two charges. - Distance between \(q_1\) and \(q_3\): \[ r_{13} = |-0.600 - (-0.400)| = 0.200 \, \text{m} \] - Distance between \(q_2\) and \(q_3\): \[ r_{23} = |0 - (-0.400)| = 0.400 \, \text{m} \]
04

Calculate Force Exerted by q1

Substitute the values into Coulomb's Law to find the force exerted by \(q_1\) on \(q_3\): \[ F_{13} = k \frac{|q_1 q_3|}{r_{13}^2} = (8.99 \times 10^9) \frac{|-1.50 \times 10^{-9} \times 5.00 \times 10^{-9}|}{(0.200)^2} \] Calculate to get: \[ F_{13} = 1.69 \times 10^{-6} \, \text{N} \] This force is attractive since \(q_1\) is negative and \(q_3\) is positive, so it acts towards \(q_1\) (downwards along the y-axis).
05

Calculate Force Exerted by q2

Now calculate the force exerted by \(q_2\) on \(q_3\): \[ F_{23} = k \frac{|q_2 q_3|}{r_{23}^2} = (8.99 \times 10^9) \frac{3.20 \times 10^{-9} \times 5.00 \times 10^{-9}}{(0.400)^2} \] Calculate to get: \[ F_{23} = 8.99 \times 10^{-7} \, \text{N} \] This force is repulsive because both \(q_2\) and \(q_3\) are positive, so it acts upwards along the y-axis.
06

Determine Net Force

To find the net force \(F_3\), add the forces vectorially along the y-axis, considering their directions:\[ F_3 = F_{23} - F_{13} \] \[ F_3 = 8.99 \times 10^{-7} \, N - 1.69 \times 10^{-6} \, N = -7.91 \times 10^{-7} \, N \] The negative sign indicates that the net force is directed downwards along the y-axis.
07

Conclusion

The magnitude of the net force on \(q_3\) is \(7.91 \times 10^{-7} \, N\) and it is directed downwards along the y-axis.

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Key Concepts

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

Electrostatics
Electrostatics is the branch of physics that deals with the study of stationary or slow-moving electric charges. It is fundamental to understanding how charges interact with each other through electric fields.

Key features of electrostatics include:
  • Charge conservation: Charge can neither be created nor destroyed; it only transfers from one place to another.
  • Coulomb's Law: This is a crucial law in electrostatics, which is used to calculate the force between two point charges. According to this law, the electrostatic force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.
  • Nature of charge: Charges can be positive or negative, and like charges repel while opposite charges attract.
Understanding electrostatics helps us explore various phenomena such as static cling, lightning, and how capacitors store electrical energy in circuits.

In problems such as the one given, electrostatics allows us to determine how charged particles influence each other through forces, which can be calculated using well-defined formulas and principles.
Point Charges
Point charges are idealized charges that are assumed to be concentrated at a single point in space. This simplification allows for easier calculation of electric forces and fields in theoretical problems. In practice, real charges could be close to point charges if the distribution of the charge is uniform or negligible in size.

When dealing with point charges in electrostatics, several factors are considered:
  • Position: The location of each charge plays a critical role in determining the forces between them, as seen through their coordinates, such as on the y-axis in this exercise.
  • Charge Sign: Positive and negative signs determine the nature of the force (attractive or repulsive).
  • Distance: The distance between point charges, often denoted as 'r', is a key variable in Coulomb's Law.
When solving problems involving point charges, these elements are used in conjunction with Coulomb's Law to determine the net force experienced by a charge due to others in its vicinity. This abstraction is particularly useful in simplifying complex electrostatic interactions.
Force Magnitude and Direction
The force magnitude and direction describe how much force is applied by one charge on another and in which direction. This is necessary to fully understand the interactions described by Coulomb's Law.

In the context of the problem:
  • Magnitude: The formula \[ F = k \frac{|q_1 q_2|}{r^2} \] is used to calculate how strong the force is. The constant \(k\) is the electrostatic constant and ensures the units are correct.
  • Direction: Determined by the signs of the charges. A negative force indicates attraction, while a positive force indicates repulsion. For example, in the scenario given, the net force direction is down since the net force calculated is negative.
Calculating these aspects allows us to predict the motion of charges. Forces in opposite directions are vectorially added to find the net effect. A thorough understanding of the force's magnitude and direction is essential for predicting the behavior of charged objects in an electrostatic field.

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Most popular questions from this chapter

Point charges \(q_{1}=-5 \mathrm{nC}\) and \(q_{2}=+5 \mathrm{nC}\) are separated by \(3 \mathrm{~mm}\), forming an electric dipole. (a) Find the electric dipole moment (magnitude and direction). (b) The charges are in a uniform electric field whose direction makes an angle of \(37^{\circ}\) with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude \(7.5 \times 10^{-9} \mathrm{~N} \cdot \mathrm{m}\) ?

A \(+2.00-n C\) point charge is at the origin, and a second \(-5.00-n C\) point charge is on the \(x\)-axis at \(x=0.800 \mathrm{~m} .\) (a) Find the electric field (magnitude and direction) at each of the following points on the \(x\)-axis: (i) \(x=0.200 \mathrm{~m}\); (ii) \(x=-0.200 \mathrm{~m}\). (b) Find the net electric force that the two charges would exert on an electron placed at each point in part (a).

Two thin rods of length \(L\) lie along the \(x\)-axis, one between \(x=a / 2\) and \(x=a / 2+L\) and the other between \(x=-a / 2\) and \(x=-a / 2-L .\) Each rod has positive charge \(Q\) distributed uniformly along its length. (a) Calculate the electric field produced by the second rod at points along the positive \(x\)-axis. (b) Show that the magnitude of the force that one rod exerts on the other is $$ F=\frac{Q^{2}}{4 \pi \epsilon_{0} L^{2}} \ln \left[\frac{(a+L)^{2}}{a(a+2 L)}\right] $$ (c) Show that if \(a \gg L, \quad\) the magnitude of this force reduces to \(F=Q^{2} / 4 \pi \epsilon_{0} a^{2} . \quad\) (Hint: Use the expansion \(\ln (1+z)=z-\) \(z^{2} / 2+z^{3} / 3-\cdots\), valid for \(|z| \ll 1\). Carry all expansions to at least order \(L^{2} / a^{2}\).) Interpret this result.

Particles in a Gold Ring. You have a pure (24-karat) gold ring with mass \(19.7 \mathrm{~g}\). Gold has an atomic mass of \(197 \mathrm{~g} / \mathrm{mol}\) and an atomic number of 79. (a) How many protons are in the ring, and what is their total positive charge? (b) If the ring carries no net charge, how many electrons are in it?

A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, \(2.25 \mathrm{~cm}\) distant from the first, in a time interval of \(1.50 \times 10^{-6}\) s. (a) Find the magnitude of the electric field. (b) Find the speed of the proton when it strikes the negatively charged plate.
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