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Chapter 21: Problem 3

(1) What is the magnitude of the force a \(+ 25 \mu \mathrm { C }\) charge exerts on a \(+ 2.5 \mathrm { mC }\) charge 28\(\mathrm { cm }\) away?

Short Answer

Expert verified
The magnitude of the force is approximately 7.17 mN.

Step by step solution

01

Identify Given Values

First, let's list the given values:- The first charge \( q_1 = 25 \mu \text{C} = 25 \times 10^{-6} \text{C} \).- The second charge \( q_2 = 2.5 \text{mC} = 2.5 \times 10^{-3} \text{C} \).- Distance between charges \( r = 28 \text{cm} = 0.28 \text{m} \).
02

Use Coulomb's Law Formula

Coulomb's Law is given by the formula:\[F = k \frac{|q_1 q_2|}{r^2}\]where \( k = 8.99 \times 10^9 \text{N} \cdot \text{m}^2/\text{C}^2 \) is the electrostatic constant.
03

Substitute Values

Substitute the given values into the Coulomb's Law equation:\[F = 8.99 \times 10^9 \frac{|25 \times 10^{-6} \times 2.5 \times 10^{-3}|}{(0.28)^2}\]
04

Perform Calculations

First, calculate the product of the charges:\[|q_1 q_2| = 25 \times 10^{-6} \times 2.5 \times 10^{-3} = 62.5 \times 10^{-9}\]Next, square the distance:\[r^2 = (0.28)^2 = 0.0784\]Now, substitute these values back into the formula:\[F = 8.99 \times 10^9 \times \frac{62.5 \times 10^{-9}}{0.0784}\]Calculate this expression:\[F = 8.99 \times 10^9 \times 797.704 = 7.17 \times 10^{-3} \text{ N}\]
05

Round the Result

The calculated force is approximately 0.00717 N. To express this in three significant figures, this becomes 0.00717 N or 7.17 mN.

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Key Concepts

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

Electrostatic Force
Electrostatic force refers to the force between charged particles. This force is described by Coulomb's Law, which helps us understand how charged objects interact. The formula for Coulomb's Law is:
  • \( F = k \frac{|q_1 q_2|}{r^2} \)
  • where \( F \) is the force, \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between them.
The electrostatic force can be either attractive or repulsive:
  • Attractive if the charges are opposite (positive and negative).
  • Repulsive if the charges are the same (both positive or both negative).
This interaction occurs due to the electric fields created by the charges, which exert a force on each other.
Coulomb's Law demonstrates the inverse square relationship, meaning that the force decreases with the square of the distance between charges. This relationship is crucial in understanding electric interactions in various fields from physics to chemistry.
Electric Charge
Electric charge is a fundamental property of matter that causes it to experience a force in an electric field. There are two types of electric charges: positive and negative.
Charged objects exert forces on each other, leading to the concepts of attraction and repulsion.
The unit of charge is the coulomb (C). However, in practical situations like this exercise, smaller units are often used:
  • Microcoulombs (\( \mu \text{C} \)), where \( 1 \mu \text{C} = 10^{-6} \text{C} \)
  • Millicoulombs (\( \text{mC} \)), where \( 1 \text{mC} = 10^{-3} \text{C} \)
Understanding electric charge is fundamental to grasping electric forces and electric fields. Depending on their signs, charges can exert repelling or attracting force, which plays a critical role in phenomena like electricity and magnetism.
The interaction between charges is the central principle behind many electronic devices and natural occurrences.
Distance Between Charges
The distance between charges directly affects the magnitude of the electrostatic force between them, as described by Coulomb's Law. The relationship is inversely proportional to the square of the distance:
  • If the distance (r) increases, the force decreases rapidly, since the force is spread over a larger area.
  • If the distance decreases, the force increases, becoming stronger as charges are nearer to each other.
This distance between charges is measured in meters (m) in the SI unit system.
In the given exercise, the distance is converted from centimeters to meters to maintain consistency in units, ensuring accuracy in calculations.
Understanding how distance influences electrostatic interactions is important in fields like physics, chemistry, and engineering. It helps in designing devices like capacitors, where the distance between plates affects their capacity to store charge. In everyday life, this concept is crucial in managing static electricity, ensuring safety and functionality in electronic appliances.

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Most popular questions from this chapter

Measurements indicate that there is an electric field surrounding the Earth. Its magnitude is about \(150 \mathrm{~N} / \mathrm{C}\) at the Earth's surface and points inward toward the Earth's center. What is the magnitude of the electric charge on the Earth? Is it positive or negative? [Hint: The electric field outside a uniformly charged sphere is the same as if all the charge were concentrated at its center.]

(III) Suppose a uniformly charged wire starts at point 0 and rises vertically along the positive \(y\) axis to a length \(\ell\) . (a) Determine the components of the electric field \(E _ { x }\) and \(E _ { y }\) at point \(( x , 0 ) .\) That is, calculate \(\vec { \mathbf { E } }\) near one end of a long wire, in the plane perpendicular to the wire. (b) If the wire extends from \(y = 0\) to \(y = \infty ,\) so that \(\ell = \infty ,\) show that \(\vec { \mathbf { E } }\) makes a \(45 ^ { \circ }\) angle to the horizontal for any \(x .\) [Hint: See Example 11 of "Electric Charge and Electric Field" andFig. \(29 . ]\)

Four equal positive point charges, each of charge \(8.0 \mu \mathrm { C } ,\) are at the corners of a square of side 9.2\(\mathrm { cm } .\) What charge should be placed at the center of the square so that all charges are at equilibrium? Is this a stable or unstable equilibrium in the plane?

(II) Compare the electric force holding the electron in orbit \(\left( r = 0.53 \times 10 ^ { - 10 } \mathrm { m } \right)\) around the proton nucleus of the hydrogen atom, with the gravitational force between the same electron and proton. What is the ratio of these two forces?

(I) A proton is released in a uniform electric field, and it experiences an electric force of \(2.18 \times 10^{-14} \mathrm{~N}\) toward the south. What are the magnitude and direction of the electric field?
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