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Chapter 21: Problem 6

Two small spheres spaced 20.0 \(\mathrm{cm}\) apart have equal charge. How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is \(4.57 \times 10^{-21} \mathrm{N} ?\)

Short Answer

Expert verified
Each sphere must have 893 excess electrons.

Step by step solution

01

Understand Coulomb鈥檚 Law

Coulomb's Law calculates the force between two charged objects. The formula is \[ F = k \frac{|q_1 q_2|}{r^2} \]where:- \( F \) is the force between the charges,- \( q_1 \) and \( q_2 \) are the charges,- \( r \) is the distance between the centers of the two spheres, - \( k \) is Coulomb's constant, \( 8.99 \times 10^9 \, \mathrm{Nm^2/C^2} \). Here, \( F = 4.57 \times 10^{-21} \, \mathrm{N} \) and \( r = 0.2 \, \mathrm{m} \). Since the spheres have equal charge, let \( q_1 = q_2 = q \).
02

Solve for Charge q

Substitute the known values into Coulomb's Law:\[4.57 \times 10^{-21} = 8.99 \times 10^9 \frac{q^2}{(0.2)^2}\]Simplify and solve for \( q^2 \):\[q^2 = \left(4.57 \times 10^{-21}\right) \cdot (0.2)^2 / 8.99 \times 10^9 \]Calculate:\[q^2 \approx 2.03 \times 10^{-31}\]Take the square root:\[q \approx \sqrt{2.03 \times 10^{-31}} \approx 1.43 \times 10^{-16} \, \mathrm{C}\]
03

Calculate Excess Electrons

The charge of one electron is \( e = 1.602 \times 10^{-19} \, \mathrm{C} \). To find the number of excess electrons \( n \), use:\[n = \frac{q}{e} = \frac{1.43 \times 10^{-16}}{1.602 \times 10^{-19}}\]Calculate:\[n \approx 8.93 \times 10^2\]Since the number of electrons must be a whole number, round \( 8.93 \times 10^2 \) to the nearest whole number: \( n = 893 \).

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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 matter that enables particles to experience a force when placed in an electromagnetic field. Every charged particle carries a certain quantity of charge, measured in Coulombs (C). There are two types of electric charges: positive and negative.

Protons possess a positive charge, while electrons carry a negative charge. The interaction of these charges is what leads to electric forces acting between objects. The unit of electric charge is derived from the charge of an electron, with one electron having a charge of approximately \( 1.602 \times 10^{-19} \text{ C} \).

In scenarios where two objects have the same type of charge, either both positive or both negative, they repel each other. Conversely, opposite charges attract. This property is essential in understanding phenomena like static electricity and electric forces.
Force of Repulsion
The force of repulsion is the push that occurs when two like-charged objects encounter each other. In this context, when two small spheres with like charges are located near each other, they tend to push away from each other due to the electric force.

Coulomb's Law is used to calculate the magnitude of this force. According to Coulomb's Law, the repulsive force \( F \) between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. The expression is given as:
  • \( F = k \frac{|q_1 q_2|}{r^2} \)
Here, \( k \) is the Coulomb's constant, \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between them.
Excess Electrons
Excess electrons on an object indicate that an object has gained additional electrons making it negatively charged. In the original exercise, we determined the number of excess electrons needed to result in a known level of repulsive force between two charged spheres.

The process involves first calculating the total charge \( q \) on each sphere using Coulomb's Law. Once \( q \) is determined, the number of excess electrons can be found by dividing the total charge by the charge of a single electron:
  • \( n = \frac{q}{e} \)
  • where \( e \) is the charge of an electron, approximately \( 1.602 \times 10^{-19} \text{ C} \).
This calculation reveals how many extra electrons are present to create the specified repulsive force.
Coulomb's Constant
Coulomb's constant \( (k) \) is a key figure used in Coulomb's Law and reflects how electric forces scale in the formula. Its value is approximately \( 8.99 \times 10^9 \text{ Nm}^2/ ext{C}^2 \).

This constant helps quantify the force between two point charges in a vacuum. It is a reflection of the interaction strength of electric charges at a specific distance in an unimpeded medium, and it plays a critical role in predicting how strongly charged objects will attract or repel each other in space.

By applying Coulomb's constant, we can determine how much repulsion or attraction force will be experienced by objects depending on their charge and distance. Understanding this allows for precise calculations in both academic and practical applications.

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Most popular questions from this chapter

The ammonia molecule \(\left(\mathrm{NH}_{3}\right)\) has a dipole moment of \(5.0 \times 10^{-30} \mathrm{C} \cdot \mathrm{m} .\) Ammonia molecules in the gas phase are placed in a a uniform electric field \(\vec{\boldsymbol{E}}\) with magnitude \(1.6 \times\) \(10^{6} \mathrm{N} / \mathrm{C} .\) (a) What is the change in electric potential energy when the dipole moment of a molecule changes its orientation with respect to \(\vec{\boldsymbol{E}}\) from parallel to perpendicular? (b) At what absolute temperature \(T\) is the average translational kinetic energy \(\frac{3}{2} k T\) of a molecule equal to the change in potential energy calculated in part (a)? (Note: Above this temperature, thermal agitation prevents the dipoles from aligning with the electric field.)

CALC 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 \(a>>L,\) the magnitude of this force reduces to \(F=Q^{2} / 4 \pi \epsilon_{0} a^{2} .\) (Hint: Use the expansion \(\ln (1+z)=z-\) \(z^{2} / 2+z^{3} / 3-\cdots,\) valid for \(|z|<<1 .\) Carry all expansions to at least order \(L^{2} / a^{2} .\) ) Interpret this result.

A charge \(q_{1}=+5.00 \mathrm{nC}\) is placed at the origin of an \(x y\) -coordinate system, and a charge \(q_{2}=-2.00 \mathrm{nC}\) is placed on the positive \(x\) -axis at \(x=4.00 \mathrm{cm} .\) (a) If a third charge \(q_{3}=\) \(+6.00 \mathrm{nC}\) is now placed at the point \(x=4.00 \mathrm{cm}, y=3.00 \mathrm{cm}\) find the \(x\) - and \(y\) -components of the total force exerted on this charge by the other two. (b) Find the magnitude and direction of this force.

Particles in a Gold Ring. You have a pure (24 karat) gold ring with mass 17.7 \(\mathrm{g}\) . Gold has 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?

CP Two positive point charges \(Q\) are held fixed on the \(x\) -axis at \(x=a\) and \(x=-a .\) A third positive point charge \(q,\) with mass \(m,\) is placed on the \(x\) -axis away from the origin at a coordinate \(x\) such that \(|x| << a\) . The charge \(q,\) which is free to move along the \(x\) -axis, is then released. (a) Find the frequency of oscillation of the charge \(q .\) (Hint: Review the definition of simple harmonic motion in Section \(14.2 .\) Use the binomial expansion \((1+z)^{n}=1+n z+n(n-1) z^{2} / 2+\cdots,\) valid for the case \(|z|<1 .\) (b) Suppose instead that the charge \(q\) were placed on the \(y\) -axis at a coordinate \(y\) such that \(|y| < < a\) , and then released. If this charge is free to move anywhere in the \(x y\) -plane, what will happen to it? Explain your answer.
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