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Chapter 24: Problem 25

If two equal point charges, each of \(1 \mathrm{C}\), were separated in air by a distance of \(1 \mathrm{~km}\), what would be the force between them?

Short Answer

Expert verified
The force between them is 8988 N.

Step by step solution

01

Understand Coulomb's Law

Coulomb's Law describes the electrostatic force between two point charges. The formula is defined as \( F = \frac{k \cdot |q_1 \cdot q_2|}{d^2} \), where \( F \) is the force between the charges, \( k \) is Coulomb's constant \( 8.988 \times 10^9 \, \mathrm{Nm}^2/\mathrm{C}^2 \), \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( d \) is the distance between the charges.
02

Substitute the Given Values

Substitute the given values into the formula: \( q_1 = 1 \, \mathrm{C} \), \( q_2 = 1 \, \mathrm{C} \), and \( d = 1 \, \mathrm{km} = 1000 \, \mathrm{m} \). Thus, \( F = \frac{8.988 \times 10^9 \cdot 1 \cdot 1}{(1000)^2} \).
03

Simplify the Expression

Now perform the calculations. First, calculate the distance squared: \( (1000)^2 = 1000000 \). Then, plug it into the formula: \( F = \frac{8.988 \times 10^9}{1000000} \).
04

Calculate the Force

Divide the values to find the force: \( F = 8.988 \times 10^3 \). This simplifies to \( F = 8988 \, \mathrm{N} \).
05

Interpret the Result

The resulting force of 8988 Newtons indicates a significant repulsive force between the two charges. This force is felt over a large distance due to the large charge magnitude.

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Key Concepts

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

Electrostatic Force
Electrostatic force is the fundamental force that acts between charged objects. It is a kind of non-contact force which operates due to the electrical charges of particles. According to Coulomb's Law, this force can either attract or repel depending on the nature of the charges involved. The fundamental principle is that like charges repel and unlike charges attract.
The forces are noticeable on a macroscopic level when the magnitude of the charges is significant, like in the exercise we discussed. This electrostatic interaction is pivotal in diverse fields such as physics, chemistry, and engineering.
Moreover, it plays a crucial role in everyday phenomena such as static electricity and the operation of electronic devices. Understanding the forces between charges helps in designing and predicting the behavior of systems involving electrical forces.
Point Charges
Point charges are idealized objects that carry an electric charge considered to be concentrated at a single point in space. In theoretical physics and exercises, point charges simplify the calculation of electrostatic forces, as they allow us to apply Coulomb's Law without worrying about the size or shape of the actual objects.
  • They enable a cleaner mathematical model to predict the behavior of charges and electric fields around them.
  • Technically, point charges are not physical real-world entities but are used as an abstraction to facilitate understanding of electrostatic interactions.
In our exercise, the point charges were each given as 1 Coulomb, a large charge amount implying strong forces generated over even considerable distances like 1 kilometer.
This concept aids in analyzing complex systems by breaking them down into simpler interactions between charge units.
Coulomb's Constant
Coulomb's constant is a proportionality factor in Coulomb's Law, enabling the calculation of electrostatic force between point charges. The value of Coulomb's constant, approximately r8.988 脳 10^9 Nm虏/C虏, is derived from the permittivity of free space. This constant helps in scaling the electrostatic force internationally so it can be compared across different scenarios and setups.
It provides the physical dimension necessary to the equation to ensure that the calculated force comes out in Newtons, the standard unit for force in the International System of Units (SI).
  • Coulomb's constant balances the mathematical relationship between charge quantities and their separation distance.
  • It is key in defining the magnitude of forces where electric charges are involved, such as in capacitors or faraday cages.
Coulomb's constant is fundamental in making electrostantial force calculations universally applicable, whether in vacuum conditions or in air as seen in our problem.

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Most popular questions from this chapter

Three point charges in vacuum are placed on the \(x\) -axis in Fig. \(24-\) 1 . Find the net force on the \(-5 \mu \mathrm{C}\) charge due to the two other charges. Because unlike charges attract, the forces on the \(-5 \mu \mathrm{C}\) charge are as shown. The magnitudes of \(\overrightarrow{\mathbf{F}}_{E 3}\) and \(\overrightarrow{\mathbf{F}}_{E 8}\) are given by Coulomb's Law: $$ \begin{array}{l} F_{E 3}=\left(9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right) \frac{\left(3.0 \times 10^{-6} \mathrm{C}\right)\left(5.0 \times 10^{-6} \mathrm{C}\right)}{(0.20 \mathrm{~m})^{2}}=3.4 \mathrm{~N} \\ F_{E 8}=\left(9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right) \frac{\left(8.0 \times 10^{-6} \mathrm{C}\right)\left(5.0 \times 10^{-6} \mathrm{C}\right)}{(0.30 \mathrm{~m})^{2}}=4.0 \mathrm{~N} \end{array} $$ Keep in mind the following: (1) Proper units (coulombs and meters) must be used. (2) Because we want only the magnitudes of the forces, we do not carry along the signs of the charges. That is, we use their absolute values. Determine if the forces are attractive or repulsive and then draw them in your diagram. Pick a direction to be positive and sum the forces. From the diagram, the resultant force on the center charge is $$ F_{E}=F_{E 8}-F_{E 3}=4.0 \mathrm{~N}-3.4 \mathrm{~N}=0.6 \mathrm{~N} $$ and it is in the \(+x\) -direction, to the right.

Two identical tiny metal balls carry charges of \(+3 \mathrm{nC}\) and \(-12 \mathrm{nC}\). They are \(3 \mathrm{~m}\) apart in vacuum. (a) Compute the force of attraction. (b) The balls are now touched together and then separated to \(3 \mathrm{~cm}\). Describe the forces on them now.

Imagine two separated tiny interacting uniformly charged spheres. What happens to the electrostatic force on each of them if the charge on one is doubled?

Two small charged spheres are placed in vacuum on the \(x\) -axis: \(+3.0 \mu \mathrm{C}\) at \(x=0\) and \(-5.0 \mu \mathrm{C}\) at \(x=40 \mathrm{~cm} .\) Where must a third charge \(q\) be placed if the force it experiences is to be zero? The situation is represented in Fig. 24-4. We know that \(q\) must be placed somewhere on the \(x\) -axis. (Why?) Suppose that \(q\) is positive. When it is placed in interval \(B C\), the two forces on it are in the same direction and cannot cancel. When it is placed to the right of \(C\), the attractive force from the \(-5 \mu \mathrm{C}\) charge is always larger than the repulsion of the \(+3.0 \mu \mathrm{C}\) charge. Therefore, the force on \(q\) cannot be zero in this region. Only in the region to the left of \(B\) can cancellation occur. (Can you show that this is also true if \(q\) is negative?) For \(q\) placed as shown, when the net force on it is zero, we have \(F_{E 3}=F_{E 5}\) and so, for distances in meters, $$ k_{0} \frac{q\left(3.0 \times 10^{-6} \mathrm{C}\right)}{d^{2}}=k_{0} \frac{q\left(5.0 \times 10^{-6} \mathrm{C}\right)}{(0.40 \mathrm{~m}+d)^{2}} $$ After canceling \(q, k_{0}\), and \(10^{-6} \mathrm{C}\) from each side, cross- multiply to obtain $$ 5 d^{2}=3.0(0.40+d)^{2} \text { or } d^{2}-1.2 d-0.24=0 $$ Using the quadratic formula, $$ d=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}=\frac{1.2 \pm \sqrt{1.44+0.96}}{2}=0.60 \pm 0.775 \mathrm{~m} $$ Two values, \(1.4 \mathrm{~m}\) and \(-0.18 \mathrm{~m}\), are therefore found for \(d\). The first is the correct one; the second gives the point in \(B C\) where the two forces have the same magnitude but do not cancel.

Two small spheres in vacuum are \(1.5 \mathrm{~m}\) apart center-to-center. They carry identical charges. Approximately how large is the charge on each if each sphere experiences a force of \(2 \mathrm{~N}\) ? The diameters of the spheres are small compared to the \(1.5 \mathrm{~m}\) separation. We may therefore approximate them as point charges. Coulomb's Law, \(F_{E}=k_{0} q_{\cdot 1} q_{\cdot 2} / r^{2}\), leads to $$ q_{.1} q_{.2}=q^{2}=\frac{F_{E} r^{2}}{k_{0}}=\frac{(2 \mathrm{~N})(1.5 \mathrm{~m})^{2}}{9 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}}=5 \times 10^{-10} \mathrm{C}^{2} $$ from which \(q=2 \times 10^{-5} \mathrm{C}\)
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