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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 \(3.33 \times 10^{-21} \mathrm{~N} ?\)

Short Answer

Expert verified
Use the calculated value of `e` from step 3, and the known charge of an electron to find the answer and write your short answer here. This value will be number of excess electrons.

Step by step solution

01

Understanding the Problem

Two equally charged spheres are `20.0 cm = 0.2 m` apart creating a repulsion force of `3.33 x 10^-21 N`. We need to find the number of excess electrons present on each of the spheres.
02

Use Coulomb's Law

Using Coulomb's Law `F = k(e^2 / r^2)`, where `F` is the net force between the spheres, `k` is Coulomb's constant \(8.99 x 10^9 N.m^2/C^2\), `e` is the magnitued of the charge on each sphere, `r` is the distance between the spheres. You can rearrange the formula to find the charge on each sphere: \[ e= \sqrt{(F*r^2) / k} \]. Substituting `F = 3.33 x 10^-21 N`, `r = 0.2 m` and `k = 8.99 x 10^9 N.m^2/C^2` to solve for `e`.
03

Calculate the Charge

Using the values for `F`, `r` and `k` from Step 2 into the formula \(e= \sqrt{(F*r^2) / k}\) to find the total charge `e` on each sphere.
04

Find the Excess Electrons

Electrons have a fundamental charge of `1.6 x 10^-19 C`. To find the number of excess electrons causing the repulsion, you can find the ratio of the total charge `e` to the charge of a single electron. \[number\ of\ electrons = e / charge\ of\ an\ electron\]. Substituting the value of `e` from the previous step and the charge of an electron to get the number of excess electrons.

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Key Concepts

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

Electrostatic Force
When we refer to the electrostatic force, we're discussing the interaction between stationary electric charges. It's a fundamental force that can both pull charges together (attractive force) and push charges apart (repulsive force). In the context of the given problem, we are dealing with the electrostatic force as a repulsive force because the two small spheres are charged equally and, as such, repel each other.

According to Coulomb鈥檚 Law, the force (\(F\)) between two point charges is directly proportional to the product of the magnitude of the charges (\(e_1e_2\text{ or }e^2\)) and inversely proportional to the square of the distance (\(r^2\text{ or }0.2^2 m^2\)) between them. The given exercise asks for the determination of excess electrons based on the repulsive force (\(3.33 \times 10^{-21}\text{N}\text{ or }0.2m^2\)) between these charges, which underscores the practical application of the electrostatic force concept.
Excess Electrons
Delving into the notion of excess electrons, we tackle the idea that objects can gain or lose electrons to obtain a net charge. When an object has more electrons than protons, it carries a negative charge. Those extra electrons that cause the object to be negatively charged are referred to as 'excess electrons.'

In our context, the excess electrons on each sphere create a repulsive force. The more excess electrons, the greater the charge and the stronger the force between the objects. To find the number of excess electrons, one needs to know the total charge and then divide it by the charge of a single electron (\(1.6 \times 10^{-19} C\text{ or }the elementary charge\)). The calculation from the step-by-step solution reveals the quantity of these excess electrons responsible for the measured electrostatic force between the spheres.
Coulomb's Constant
An essential component in electrostatics is Coulomb's constant (\(k\text{ or }8.99 \times 10^9 \frac{N\text{m}^2}{C^2}\)), which facilitates the computation of electrostatic force between two point charges as part of Coulomb's Law. This constant helps us quantify the electrostatic force and serves as a bridge between the conceptual and the quantitative aspects of assessing charges.

Coulomb's constant arises from the equation \text{F} = k\frac{\text{e}^2}{\text{r}^2}\text{, directly linking the magnitude of the force with the charges and the distance between them}. In the exercise, the value of Coulomb's constant is crucial for determining the charge on each sphere. It underscores the universality of electrostatic interactions, providing a way to calculate the force across various conditions regardless of the specific charges or distances 鈥� as long as they are point charges and at rest. It鈥檚 truly a cornerstone of the calculation that allows us to solve for the unknown quantities.

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Most popular questions from this chapter

A nerve signal is transmitted through a neuron when an excess of \(\mathrm{Na}^{+}\) ions suddenly enters the axon, a long cylindrical part of the neuron. Axons are approximately \(10.0 \mu \mathrm{m}\) in diameter, and measurements show that about \(5.6 \times 10^{11} \mathrm{Na}^{+}\) ions per meter (each of charge \(+e\) ) enter during this process. Although the axon is a long cylinder, the charge does not all enter everywhere at the same time. A plausible model would be a series of point charges moving along the axon. Consider a \(0.10 \mathrm{~mm}\) length of the axon and model it as a point charge. (a) If the charge that enters each meter of the axon gets distributed uniformly along it, how many coulombs of charge enter a \(0.10 \mathrm{~mm}\) length of the axon? (b) What electric field (magnitude and direction) does the sudden influx of charge produce at the surface of the body if the axon is \(5.00 \mathrm{~cm}\) below the skin? (c) Certain shark can respond to electric fields as weak as \(1.0 \mu \mathrm{N} / \mathrm{C}\). How far from this segment of axon could a shark be and still detect its electric field?

In a follow-up experiment, a charge of \(+40 \mathrm{pC}\) was placed at the center of an artificial flower at the end of a \(30-\mathrm{cm}\) -long stem. Bees were observed to approach no closer than \(15 \mathrm{~cm}\) from the center of this flower before they flew away. This observation suggests that the smallest external electric field to which bees may be sensitive is closest to which of these values? (a) \(2.4 \mathrm{~N} / \mathrm{C} ;\) (b) \(16 \mathrm{~N} / \mathrm{C} ;\) (c) \(2.7 \times 10^{-10} \mathrm{~N} / \mathrm{C}\) (d) \(4.8 \times 10^{-10} \mathrm{~N} / \mathrm{C}\).

A small \(12.3 \mathrm{~g}\) plastic ball is tied to a very light \(28.6 \mathrm{~cm}\) string that is attached to the vertical wall of a room (Fig. \(\mathbf{P} 2 \mathbf{1 . 6 3}\) ). A uniform horizontal electric field exists in this room. When the ball has been given an excess charge of \(-1.11 \mu \mathrm{C},\) you observe that it remains suspended, with the string making an angle of \(17.4^{\circ}\) with the wall. Find the magnitude and direction of the electric field in the room.

Two \(1.20 \mathrm{~m}\) nonconducting rods meet at a right angle. One rod carries \(+2.50 \mu \mathrm{C}\) of charge distributed uniformly along its length, and the other carries \(-2.50 \mu \mathrm{C}\) distributed uniformly along it (Fig. \(\mathbf{P} 2 \mathbf{1 . 8 5}\) ). (a) Find the magnitude and direction of the electric field these rods produce at point \(P,\) which is \(60.0 \mathrm{~cm}\) from each rod. (b) If an electron is released at \(P\), what are the magnitude and direction of the net force that these rods exert on it?

(a) What must the charge (sign and magnitude) of a \(1.45 \mathrm{~g}\) particle be for it to remain stationary when placed in a downwarddirected electric field of magnitude \(650 \mathrm{~N} / \mathrm{C} ?\) (b) What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight?
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