Ch 21: Electric Charge and Electric Field

Young & Freedman Calc - University Physics 14th Edition

Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook

Young & Freedman Calc 14th Edition

Ch 21: Electric Charge and Electric Field

Problem 9b

Chapter 21, Problem 9b

Two small plastic spheres are given positive electric charges. When they are $15.0$ cm apart, the repulsive force between them has magnitude $0.220$ N. What is the charge on each sphere if one sphere has four times the charge of the other?

Verified step by step guidance

Start by understanding Coulomb's Law, which describes the force between two point charges. The formula is:

F = \frac{k q Q}{r^{2}}

, where

F

is the force,

k

is Coulomb's constant (

8.99 × 10 9^{} N m^{2} / C^{2}

q

and

Q

are the charges, and

r

is the distance between the charges.

Convert the distance from centimeters to meters, as the standard unit for distance in physics is meters. So,

15.0 cm = 0.15 m

Since one sphere has four times the charge of the other, let the charge on the smaller sphere be

q

and the charge on the larger sphere be

4 q

. Substitute these into Coulomb's Law:

F = \frac{k q 4 q}{r^{2}}

Rearrange the equation to solve for

q

q = \sqrt{\frac{F r^{2}}{4 k}}

. Substitute the known values for

F

r

, and

k

into the equation.

Once you have calculated

q

, multiply it by 4 to find the charge on the larger sphere. This will give you the charge values for both spheres.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Coulomb's Law

Coulomb's Law describes the electrostatic interaction between charged particles. It states that the force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. The formula is F = k * |q1 * q2| / r^2, where k is the Coulomb's constant.

Electric Charge

Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electromagnetic field. Charges can be positive or negative, and like charges repel while opposite charges attract. In this problem, understanding the distribution of charge between the spheres is crucial for calculating the force.

Proportional Relationships

Proportional relationships involve the comparison of two quantities where one is a constant multiple of the other. In this scenario, one sphere has four times the charge of the other, which means if the charge of one sphere is q, the other is 4q. This relationship is key to solving for the individual charges using Coulomb's Law.

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Related Practice

Textbook Question

Suppose you had two small boxes, each containing $1.0$ g of protons.

(a) If one were placed on the moon by an astronaut and the other were left on the earth, and if they were connected by a very light (and very long!) string, what would be the tension in the string? Express your answer in newtons and in pounds. Do you need to take into account the gravitational forces of the earth and moon on the protons? Why?

(b) What gravitational force would each box of protons exert on the other box?

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Textbook Question

Two small aluminum spheres, each having mass $0.0250$ kg, are separated by $80.0$ cm. What fraction of all the electrons in each sphere does this represent?

2009

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Textbook Question

Three point charges are arranged on a line. Charge $q sub 3 equals positive 5.00$ nC and is at the origin. Charge $q_{2} = - 3.00$ nC and is at $x equals positive 4.00$ cm. Charge $q sub 1$ is at $x equals positive 2.00$ cm. What is $q sub 1$ (magnitude and sign) if the net force on $q sub 3$ is zero?

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Textbook Question

Two small plastic spheres are given positive electric charges. When they are $15.0$ cm apart, the repulsive force between them has magnitude $0.220$ N. What is the charge on each sphere if the two charges are equal?

2071

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Textbook Question

Two small aluminum spheres, each having mass $0.0250$ kg, are separated by $80.0$ cm. How many electrons would have to be removed from one sphere and added to the other to cause an attractive force between the spheres of magnitude $1.00 \times 10^{4}$ N (roughly $1$ ton)? Assume that the spheres may be treated as point charges.

The nuclei of large atoms, such as uranium, with $92$ protons, can be modeled as spherically symmetric spheres of charge. The radius of the uranium nucleus is approximately $7.4$\times$10^{-15}$ m. What magnitude of electric field does it produce at the distance of the electrons, which is about $1.0 \times 10^{- 10}$ m?

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Two small plastic spheres are given positive electric charges. When they are 15.015.015.0 cm apart, the repulsive force between them has magnitude 0.2200.2200.220 N. What is the charge on each sphere if one sphere has four times the charge of the other?

Verified video answer for a similar problem:

Key Concepts

Coulomb's Law

Electric Charge

Proportional Relationships

Two small plastic spheres are given positive electric charges. When they are $15.0$ cm apart, the repulsive force between them has magnitude $0.220$ N. What is the charge on each sphere if one sphere has four times the charge of the other?