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Chapter 21: Problem 22

Two positive point charges \(q\) are placed on the \(y\) -axis at \(y=a\) and \(y=-a .\) A negative point charge \(-Q\) is located at some point on the \(+x\) -axis. (a) In a free-body diagram, show the forces that act on the charge \(-Q .\) (b) Find the \(x\) -and \(y\) -components of the net force that the two positive charges exert on \(-Q\) . (Your answer \(r\) . should involve only \(k, q, Q, a\) and the coordinate \(x\) of the third charge.) (c) What is the net force on the charge \(-Q\) when it is at the origin \((x=0) ?(\mathrm{d})\) Graph the \(x\) -component of the net force on the charge \(-Q\) as a function of \(x\) for values of \(x\) between \(-4 a\) and \(+4 a\) .

Short Answer

Expert verified
The net force on \\(-Q\\) at the origin is zero. The \\x\\-component depends symmetrically on \\x\\.

Step by step solution

01

Analyze Forces on Charge -Q

The negative charge \(-Q\) experiences forces due to both positive charges \(+q\). Since these charges are symmetrically placed on the \(+y\) and \(-y\) axes, forces will be exerted along lines connecting the charges with \(-Q\). For both charges, the force will have a component along the \(-x\)-axis and opposite \(+y\) or \(-y\)-axes, depending on their position.
02

Calculate Distance (r)

For each positive charge \(+q\), the distance to the negative charge \(-Q\) is calculated. By Pythagoras' theorem, the distance \r\ between \(-Q\) at \(x, 0\) and \(+q\) at \(0, \pm a\) is: \[r = \sqrt{x^2 + a^2}.\]
03

Determine Force from a Single Charge (F)

Using Coulomb's Law, compute the force exerted by one \(+q\) charge on \(-Q\): \[F = \frac{kqQ}{r^2} = \frac{kqQ}{x^2 + a^2}.\]
04

Find x-Component of Force (F_x)

Using trigonometry, find the \x\-component of this force from symmetry: \\[F_{x} = F \cdot \frac{x}{r} = \frac{kqQ \cdot x}{(x^2 + a^2)^{3/2}}.\] This is the same for both charges due to symmetry.
05

Find y-Component of Force (F_y)

Similarly, the \y\-component of force \F_{y}\ for one charge is: \[F_{y} = F \cdot \frac{a}{r} = \frac{kqQ \cdot a}{(x^2 + a^2)^{3/2}}.\] Each exerts opposite and equal magnitudes, thus canceling out net \y\-components.
06

Calculate Net Force at Origin (x=0)

At \(x=0\), \\[F_{net} = 2 \times \frac{kqQ \cdot 0}{a^3} = 0.\] The total force is zero due to symmetry.
07

Graph x-Component of Net Force

Plot \F_x\ as a function of \x\ ranging from \-4a\ to \+4a\. The formula for \F_x\ is: \[F_x = 2 \times \frac{kqQ \cdot x}{(x^2 + a^2)^{3/2}}.\] Evaluate and draw the function, ensuring both side symmetries and zero crossing at \(x=0\).

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Key Concepts

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

Coulomb's Law
Did you know that electricity can exert a force without touching? This phenomenon is known as electrostatic force. And it's all explained by Coulomb's Law. In simple terms, Coulomb's Law tells us how to calculate the force between two point charges.
The formula for this is: \[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \] Where:
  • \( F \) is the magnitude of the force between the charges,
  • \( k \) is Coulomb's constant, approximately equal to \( 8.99 \times 10^9 \) \( N \cdot m^2/C^2 \),
  • \( q_1 \) and \( q_2 \) are the amounts of each charge,
  • \( r \) is the distance between the centers of the two charges.
According to this law, like charges repel and opposite charges attract. So, if one charge is positive and the other is negative, they will pull toward each other. If both are negative or both are positive, they push away. This principle is crucial for predicting how charged particles interact.
Point Charges
In the world of electrostatics, we often talk about point charges. These are idealized charges that are considered to be concentrated at a single point in space.
Because point charges are modeled as having no size, they are extremely useful for simplifying calculations related to electric forces.
How do we use point charges in electrostatic problems? Let's break it down.
  • We assume the charge is perfectly spherical and infinitesimally small.
  • We can then treat it as a point source of the electric field.
  • This allows us to apply Coulomb's Law effectively.
For example, in our exercise, the charges are positioned on the \( y \)-axis, and we treat them as point charges. This way, we can use symmetry and the distances calculated from their positions to find forces acting in the system. The simplicity of point charges makes complex problems much more approachable.
Net Force Calculation
When multiple forces act on an object, like our negative charge \(-Q\), we need to determine the net force to understand the overall effect. This involves calculating both the magnitude and direction of all individual forces and then summing them up vectorially.
In this case:
  • The horizontal or \( x \)-component of the force from each positive charge is calculated using \[ F_{x} = \frac{kqQ \cdot x}{(x^2 + a^2)^{3/2}} \] and because both \( +q \) charges pull with equal and opposite horizontal components, the net force is simply their sum.
  • For the vertical \( y \)-components, they are opposed and equal. Hence they cancel each other out, making net \( F_y \) zero.
Summing these vectors gives you the net force experienced by \(-Q\), showing that careful analysis of each component is necessary. At the origin, symmetry ensures the forces from both positive charges cancel perfectly, resulting in no net force.
Symmetry in Physics
Symmetry can make solving physics problems easier. It helps us predict and calculate forces and other attributes.
In the given exercise, both positive charges are placed symmetrically around the origin on the \( y \)-axis.
Let's see how symmetry aids our solution.
  • Because the forces acting on \(-Q\) from both positive charges are equidistant and in opposite directions, they create equal but opposite vertical force components.
  • This symmetry ensures that these vertical components cancel each other out, leaving no net force in the \( y \)-direction.
  • In the horizontal direction, since both forces pull in the \(-x\)-direction, they add up rather than cancel out.
Utilizing symmetry allowed us to simplify the problem significantly. We derived that the net force would always be along the \( x \)-axis except when the negative charge is at the origin, further simplifying computations.

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Most popular questions from this chapter

A point change is at the origin. With this point charge as the source point, what is the unit vector \(\hat{r}\) in the direction of \((a)\) the field point at \(x=0, y=-1.35 \mathrm{m} ;(b)\) the field point at \(x=\) \(12.0 \mathrm{cm}, y=12.0 \mathrm{cm} ;(c)\) the field point at \(x=-1.10 \mathrm{m}, y=\) 2.60 \(\mathrm{m} ?\) Express your results in terms of the unit vectors \(\hat{i}\) and \(\hat{j}\) .

Lightning occurs when there is a flow of electric charge (principally electrons) between the ground and a thundercloud. The maximum rate of charge flow in a lightning bolt is about \(20,000 \mathrm{C} / \mathrm{s} ;\) this lasts for 100\(\mu \mathrm{s}\) or less. How much charge flows between the ground and the cloud in this time? How many electrons flow during this time?

A small sphere with mass \(m\) carries a positive charge \(q\) and is attached to one end of a silk fiber of length \(L\) . The other end of the fiber is attached to a large vertical insulating sheet that has a positive surface charge density \(\sigma\) . Show that when the sphere is in equilibrium, the fiber makes an angle equal to arctan \(\left(q \sigma / 2 m g \epsilon_{0}\right)\) with the vertical shee..

A Parallel Universe. Imagine a parallel universe in which the electric force has the same properties as in our universe but there is no gravity. In this parallel universe, the sun carries charge \(Q\) , the earth carries charge \(-Q\) , and the electric attraction between them keeps the earth in orbit. The earth in the parallel universe has the same mass, the same orbital radius, and the same orbital period as in our universe. Calculate the value of \(Q\) . (Consult Appendix F as needed)

Two small aluminum spheres, each having mass 0.0250 \(\mathrm{kg}\) are separated by \(80.0 \mathrm{cm} .\) (a) How many electrons does each sphere contain? (The atomic mass of aluminum is 26.982 \(\mathrm{g} / \mathrm{mol}\) , and its atomic number is \(13 . )\) (b) 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 \(1.00 \times 10^{4} \mathrm{N}\) (roughly 1 ton \() ?\) Assume that the spheres may be treated as point charges. (c) What fraction of all the electrons in each sphere does this represent?
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