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Chapter 21: Problem 12

A negative charge \(-0.550 \mu C\) exerts an upward \(0.200-\mathrm{N}\) force on an unknown charge 0.300 \(\mathrm{m}\) directly below it. (a) What is the unknown charge (magnitude and sign)? (b) What are the magnitude and direction of the force that the unknown charge exerts on the \(-0.550-\mu \mathrm{C}\) charge?

Short Answer

Expert verified
(a) The unknown charge is +3.65 渭C. (b) The force is 0.200 N downward.

Step by step solution

01

Understand Coulomb's Law

Coulomb's law states that the force between two point charges is given by \( F = k \frac{|q_1 q_2|}{r^2} \), where \( F \) is the force, \( k = 8.988 \times 10^9 \, \mathrm{N\,m^2/C^2} \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between the charges.
02

Plug Known Values into Coulomb's Law

We know \( F = 0.200 \, \mathrm{N} \), \( q_1 = -0.550 \, \mu \mathrm{C} = -0.550 \times 10^{-6} \, \mathrm{C} \) and \( r = 0.300 \, \mathrm{m} \). Plug these values into Coulomb's law:\[ 0.200 = 8.988 \times 10^9 \frac{|-0.550 \times 10^{-6} \cdot q_2|}{0.300^2} \]
03

Solve for Unknown Charge Magnitude

Rearrange the equation to solve for \( q_2 \):\[ q_2 = \frac{0.200 \times 0.300^2}{8.988 \times 10^9 \times 0.550 \times 10^{-6}} \]Calculate \( q_2 \):\[ q_2 = \frac{0.200 \times 0.090}{8.988 \times 10^9 \times 0.550 \times 10^{-6}} \approx 3.65 \times 10^{-6} \, \mathrm{C} \]
04

Determine the Sign of the Unknown Charge

Since a negative charge exerts an upward force on an unknown charge below it, and the force is attractive (between opposite signs), the unknown charge must be positive. Therefore, \( q_2 = +3.65 \times 10^{-6} \, \mathrm{C} \).
05

Determine the Force Exerted by the Unknown Charge

According to Newton's third law, the force exerted by the unknown charge on the \(-0.550 \, \mu \mathrm{C} \) charge will have the same magnitude but opposite direction as the force exerted by the \(-0.550 \, \mu \mathrm{C} \) charge. Thus, the force magnitude is 0.200 N and the direction is downward, towards the \(-0.550 \, \mu \mathrm{C} \) charge.

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Key Concepts

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

Electric Force
Electric force is a fundamental interaction between charged objects. It's a type of force exerted by charged particles on one another. This force can either attract or repel based on the nature of the charges involved. An important equation, known as Coulomb's Law, helps calculate this force.Coulomb's Law gives us:\[ F = k \frac{|q_1 q_2|}{r^2} \]where:
  • \( F \) stands for the force between the charges,
  • \( k \) is the Coulomb's constant, approximately \( 8.988 \times 10^9 \, \mathrm{N\,m^2/C^2} \),
  • \( q_1 \) and \( q_2 \) are the charges in question,
  • \( r \) is the distance separating these charges.
Under this law, the electric force decreases as the distance between charges increases. This is akin to the inverse square law observed in gravity.
Point Charges
Point charges are idealized charges that are thought of as existing at a single point in space. In the context of Coulomb's Law, these charges are assumed to have no physical size, allowing calculations to be made without considering the complexities of spatial dimensions.In many physics problems, we treat charges as point charges to simplify calculations and focus on the interaction itself rather than the physical properties of the charge holder.In the given exercise:
  • The \(-0.550 \, \mu \mathrm{C} \) charge is considered a point charge exerting an upward force.
  • Another charge lies directly below it at a known distance, further simplifying the setup as they can be modeled in a straight vertical line, making calculations straightforward.
This simplification allows us to use Coulomb's Law effectively as it assumes point charge interactions.
Charge Interaction
Charge interaction refers to the behavior and effect charges have on each other, particularly in terms of attraction or repulsion. Charges are generally classified into two types: positive and negative.Here are the rules for charge interaction:
  • Like charges (both positive or both negative) repel each other.
  • Opposite charges (one positive, one negative) attract each other.
In the problem, the given charge \(-0.550 \, \mu \mathrm{C} \) is pulling the unknown charge upward, indicating an attractive force.Since attraction occurs between opposite charges, the unknown charge must be positive. This conclusion is critical for further analysis, including solving for unknowns using Coulomb's Law.
Newton's Third Law
Newton's third law is often phrased as "for every action, there is an equal and opposite reaction." This principle is vital in understanding mutual force interactions between charges.In the context of electric forces:
  • If Charge A exerts a force on Charge B, Charge B exerts an equal and opposite force on Charge A.
In the exercise, if the known \(-0.550 \, \mu \mathrm{C} \) charge exerts a force of 0.200 N upwards on the unknown charge, then according to Newton's third law, the unknown charge must exert a 0.200 N force back on the \(-0.550 \, \mu \mathrm{C} \) charge, but in the opposite direction (downwards).This reflection of force helps ensure consistency across calculations and reinforces the nature of equal and opposite interactions found in physics.

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Most popular questions from this chapter

A straight, nonconducting plastic wire 8.50 \(\mathrm{cm}\) long carries a charge density of \(+175 \mathrm{nC} / \mathrm{m}\) distributed uniformly along its length. It is lying on a horizontal tabletop. (a) Find the magnitude and direction of the electric field this wire produces at a point 6.00 \(\mathrm{cm}\) directly above its midpoint. (b) If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point 6.00 \(\mathrm{cm}\) directly above its center.

A charge of \(-6.50 \mathrm{nC}\) is spread uniformly over the surface of one face of a nonconducting disk of radius \(1.25 \mathrm{cm} .\) (a) Find the magnitude and direction of the electric field this disk produces at a point \(P\) on the axis of the disk a distance of 2.00 \(\mathrm{cm}\) from its center. (b) Suppose that the charge were all pushed away from the center and distributed uniformly on the outer rim of the disk. Find the magnitude and direction of the electric field at point \(P .\) (c) If the charge is all brought to the center of the disk, find the magnitude and direction of the electric field at point \(P .\) (d) Why is the field in part (a) stronger than the field in part (b)? Why is the field in part (c) the strongest of the three fields?

(a) What is the electric field of an iron nucleus at a distance of \(6.00 \times 10^{-10} \mathrm{m}\) from the nucleus? The atomic number of iron is \(26 .\) Assume that the nucleus may be treated as a point charge. (b) What is the electric field of a proton at a distance of \(5.29 \times 10^{-11} \mathrm{m}\) from the proton? (This is the radius of the electron orbit in the Bohr model for the ground state of the hydrogen atom.)

Sketch the electric field lines for a disk of radius \(R\) with a positive uniform surface charge density \(\sigma\) . Use what you know about the electric field very close to the disk and very far from the disk to make your sketch.

A positive point charge \(q\) is placed on the \(+y\) -axis at \(y=a\) and a negative point clarge \(-q\) is placed on the \(-y\) -axis at \(y=-a\) . 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 charges \(q\) and \(-q\) exert on \(-Q .\) (Your answer 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 \((x=0)\) ? (d) Graph the \(y\) -component of the net force on the charge \(-Q\) as a function of \(x\) for values of \(x\) between \(-4 a\) and \(+4 a\) .
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