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Chapter 16: Problem 10

Two small metal spheres are \(25.0 \mathrm{cm}\) apart. The spheres have equal amounts of negative charge and repel each other with a force of \(0.036 \mathrm{N}\). What is the charge on each sphere?

Short Answer

Expert verified
Each sphere has a charge of approximately \( 1.5 \times 10^{-6} \mathrm{C} \).

Step by step solution

01

Identify the Known Values

We are given the force of repulsion between the two spheres as \( F = 0.036 \mathrm{N} \) and the distance separating them as \( r = 25.0 \mathrm{cm} = 0.25 \mathrm{m} \). We need to calculate the charge on each sphere.
02

Understand Coulomb's Law

Coulomb's law states that the force \( F \) between two point charges \( q_1 \) and \( q_2 \) separated by distance \( r \) is given by \[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \] where \( k \) is Coulomb's constant \( 8.99 \times 10^9 \mathrm{N\cdot m^2/C^2} \). For this scenario, \( q_1 = q_2 = q \).
03

Set Up the Equation Using Coulomb's Law

Substitute the known values into Coulomb's law equation: \( 0.036 = \frac{8.99 \times 10^9 \times q^2}{(0.25)^2} \). This relates the force, distance, and charge.
04

Solve for the Charge \( q \)

Rearrange the equation to solve for \( q^2 \): \( q^2 = \frac{0.036 \times 0.25^2}{8.99 \times 10^9} \). Calculate the value of \( q^2 \).
05

Calculate the Charge \( q \)

Taking the square root of the result from the previous step gives \( q \). Compute \( q \) to find the charge on each sphere: \[ q = \sqrt{\frac{0.036 \times 0.0625}{8.99 \times 10^9}} \]. Simplify and compute this to get the final answer.

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Key Concepts

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

Electric Charge
Electric charge is a fundamental property of some subatomic particles, like electrons and protons. These charges cause them to experience attraction or repulsion forces.
A charge can be either positive or negative. Electrons carry a negative charge, while protons have a positive charge.
Objects usually have a balanced amount of protons and electrons, making them neutral. However, if they gain or lose electrons, they become charged.
  • An object with more electrons than protons has a negative charge.
  • An object with more protons than electrons has a positive charge.
The unit of electric charge is the coulomb (C). In our specific problem, we have metal spheres with equal negative charges, which means they will repel each other. This repulsion is due to like charges pushing away from each other.
Force of Repulsion
Force of repulsion is a push that occurs between objects with like charges. According to Coulomb's Law, this force is a result of two charged objects exerting a force on each other.
In our exercise, both spheres carry the same negative charge. This makes them repel each other with a known force of 0.036 N.
The strength of this repulsive force depends on:
  • The amount of charge on each object.
  • The distance between the charged objects.
A greater charge or shorter distance results in a stronger repulsive force. In this problem, the force of repulsion assists us in calculating the amount of charge on each sphere.
Coulomb's Constant
Coulomb's constant, denoted as \( k \), is a value used in calculating the electric force between two charges in Coulomb's Law.
Its value is approximately \( 8.99 \times 10^9 \, \mathrm{N\cdot m^2/C^2} \). This constant ties together the relationship between force, the charges involved, and the distance separating them.
In the equation:
  • \( F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \)
\( k \) helps us determine how strong the electric force is based on the magnitude of charges and their distance. Coulomb's constant ensures that our calculations are consistent with the observed forces in nature. Using this value, we apply it to the problem to solve for the charge on each sphere.

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Most popular questions from this chapter

A charge of \(63.0 \mathrm{nC}\) is located at a distance of \(3.40 \mathrm{cm}\) from a charge of \(-47.0 \mathrm{nC} .\) What are the \(x\) - and \(y\) -components of the electric field at a point \(P\) that is directly above the 63.0 -nC charge at a distance of \(1.40 \mathrm{cm} ?\) Point \(P\) and the two charges are on the vertices of a right triangle.
Two electric charges, \(q_{1}=+20.0 \mathrm{nC}\) and \(q_{2}=+10.0 \mathrm{nC}\) are located on the \(x\) -axis at \(x=0 \mathrm{m}\) and \(x=1.00 \mathrm{m}\) respectively. What is the magnitude of the electric field at the point $x=0.50 \mathrm{m}, y=0.50 \mathrm{m} ?$
A positively charged rod is brought near two uncharged conducting spheres of the same size that are initially touching each other (diagram a). The spheres are moved apart and then the charged rod is removed (diagram b). (a) What is the sign of the net charge on sphere 1 in diagram b? (b) In comparison with the charge on sphere \(1,\) how much and what sign of charge is on sphere \(2 ?\)
Three equal charges are placed on three corners of a square. If the force that \(Q_{\mathrm{a}}\) exerts on \(Q_{\mathrm{b}}\) has magnitude \(F_{\mathrm{ba}}\) and the force that \(Q_{\mathrm{a}}\) exerts on \(Q_{\mathrm{c}}\) has magnitude \(F_{\mathrm{ca}},\) what is the ratio of \(F_{\mathrm{ca}}\) to $F_{\mathrm{ba}} ?$
A dipole consists of two equal and opposite point charges \((\pm q)\) separated by a distance \(d .\) (a) Write an expression for the electric field at a point \((0, y)\) on the dipole axis. Specify the direction of the field in all four regions: \(y>\frac{1}{2} d, 0<y<\frac{1}{2} d,-\frac{1}{2} d<y<0,\) and \(y<-\frac{1}{2} d .\) (b) At distant points \((|y| \gg d),\) write a simpler, approximate expression for the field. To what power of \(y\) is the field proportional? Does this conflict with Coulomb's law? [Hint: Use the binomial approximation $\left.(1 \pm x)^{n} \approx 1 \pm n x \text { for } x \ll 1 .\right]$
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