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Chapter 19: Problem 14

Location \(A\) is \(3.00 \mathrm{~m}\) to the right of a point charge \(q .\) Location \(B\) lies on the same line and is \(4.00 \mathrm{~m}\) to the right of the charge. The potential difference between the two locations is \(V_{B}-V_{A}=45.0 \mathrm{~V}\). What is the magnitude and sign of the charge?

Short Answer

Expert verified
The charge is \(-6.10 \times 10^{-10} \, \mathrm{C}\). It is negative.

Step by step solution

01

Understand the Problem

We are given two points, A and B, placed at distances of 3.00 m and 4.00 m from a point charge q respectively, and the potential difference between these two points is \(V_{B} - V_{A} = 45.0\, \mathrm{V}\). We need to find the magnitude and sign of charge q.
02

Recall the Potential Due to a Point Charge

The electric potential due to a point charge at a distance \(r\) from the charge is given by \(V = \frac{kq}{r}\), where \(k = 8.99 \times 10^9 \, \mathrm{N}\,\mathrm{m}^2/\mathrm{C}^2\) is the Coulomb's constant and \(q\) is the charge.
03

Write the Expression for Potential Difference

The potential difference \(V_{B} - V_{A}\) can be expressed using the potentials at points B and A: \[V_{B} - V_{A} = \frac{kq}{r_B} - \frac{kq}{r_A}\] where \(r_B = 4.00\, \mathrm{m}\) and \(r_A = 3.00\, \mathrm{m}\).
04

Substitute Known Values

Substitute \(V_{B} - V_{A} = 45.0 \, \mathrm{V}\), \(r_B = 4.00\, \mathrm{m}\), and \(r_A = 3.00\, \mathrm{m}\) into the potential difference formula:\[45.0 = kq \left(\frac{1}{4.00} - \frac{1}{3.00}\right)\]
05

Simplify the Expression

Calculate \(\frac{1}{4.00} - \frac{1}{3.00}\):\[\frac{1}{4.00} = 0.25, \quad \frac{1}{3.00} \approx 0.333\]Thus, \[\frac{1}{4.00} - \frac{1}{3.00} = 0.25 - 0.333 = -0.083\]
06

Solve for the Charge q

Plug the simplified expression into the equation:\[45.0 = kq (-0.083)\]Rearrange to solve for \(q\):\[q = \frac{45.0}{k(-0.083)}\]Calculate using \(k = 8.99 \times 10^9 \, \mathrm{N}\,\mathrm{m}^2/\mathrm{C}^2\):\[q = \frac{45.0}{(-0.083) \times 8.99 \times 10^9}\]\[q \approx -6.10 \times 10^{-10} \, \mathrm{C}\]
07

Interpret the Result

The negative sign indicates that the charge is negative, and the magnitude of the charge is \(6.10 \times 10^{-10} \, \mathrm{C}\).

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Key Concepts

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

Electric Potential Difference
Electric potential difference is an essential concept in electrostatics. It describes the work done in moving a unit positive charge from one point to another in an electric field. Essentially, it is the difference in electric potential energy between two points per unit charge.

In our exercise, points A and B have different electric potentials due to their varying distances from the point charge q. The potential difference between these points, given as \(V_B - V_A = 45.0 \, \text{V}\), tells us how much electric potential energy difference is present per unit charge between these two points.

Electric potential difference is critical when analyzing the energy changes in systems involving electric fields. It's measured in volts \( (\text{V}) \), where 1 volt is equal to 1 joule per coulomb \((1 \, \text{V} = 1 \, \text{J/C})\). This measurement helps us understand the capability of an electric field to do work.
Coulomb's Law
Coulomb's Law is a fundamental principle used to describe the force between two charges. It states that the force is directly proportional to the product of the absolute values of the charges and inversely proportional to the square of the distance between them.

The formula is given as:
  • \( F = k \frac{|q_1 q_2|}{r^2} \)
where:
  • \(F\) is the force between the charges.
  • \(k\) is Coulomb's constant, approximately \(8.99 \times 10^9 \, \text{N}\,\text{m}^2/\text{C}^2\).
  • \(q_1\) and \(q_2\) are the charges.
  • \(r\) is the distance between the charges.
Coulomb's Law applies specifically to point charges and helps predict the magnitude and direction of the force experienced by each charge due to the other. Although the exercise focuses on potential difference, understanding Coulomb's Law is crucial as it underpins the calculation of forces and potential energies in electrostatic scenarios.
Point Charge
A point charge is an idealized model used in electrostatics, where the charge is considered to be concentrated at a single point in space. This concept simplifies the analysis of electric fields and potentials since it allows for easier calculations.

In our exercise, the point charge affects the electric potential at specific locations A and B. The potential due to a point charge is calculated using the formula:
  • \( V = \frac{kq}{r} \)
where:
  • \(V\) is the electric potential at a distance \(r\) from the charge.
  • \(k\) is Coulomb's constant.
  • \(q\) is the charge of the point.
  • \(r\) is the radial distance from the charge to the point of interest.
The point charge model is beneficial for theoretical studies of electrostatics. For practical applications with extended charge distributions, other models and more complex calculations may be needed. However, the point charge provides a foundational understanding of how charges interact with their surroundings through electric fields and potential differences.

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Most popular questions from this chapter

The inner and outer surfaces of a cell membrane carry a negative and positive charge, respectively. Because of these charges, a potential difference of about \(0.070 \mathrm{~V}\) exists across the membrane. The thickness of the cell membrane is \(8.0 \times 10^{-9} \mathrm{~m}\). What is the magnitude of the electric field in the membrane?

Two capacitors have the same plate separation. However, one has square plates, while the other has circular plates. The square plates are a length \(L\) on each side, and the diameter of the circular plates is \(L\). (a) If the same dielectric material were between the plates in each capacitor, which one would have the greater capacitance? (b) By putting different dielectric materials between the capacitor plates, we can make the two capacitors have the same capacitance. Which capacitor should contain the dielectric material with the greater dielectric constant? Give your reasoning in each case. The capacitors have the same capacitance because they contain different dielectric materials. The dielectric constant of the material between the square plates has a value of \(\kappa_{\text {square }}=3.00\). What is the dielectric constant \(\kappa_{\text {circle }}\) of the material between the circular plates? Be sure that your answer is consistent with your answers to the Concept Questions.

Charges \(q_{1}\) and \(q_{2}\) are fixed in place, \(q_{2}\) being located at a distance \(d\) to the right of \(\mathrm{q}_{1} . \mathrm{A}\) third charge \(q_{3}\) is then fixed to the line joining \(q_{1}\) and \(q_{2}\) at a distance \(d\) to the right of \(q_{2}\). The third charge is chosen so the potential energy of the group is zero; that is, the potential energy has the same value as that of the three charges when they are widely separated. Determine the value for \(q_{3}\), assuming that (a) \(q_{1}=q_{2}=q\) and (b) \(q_{1}=q\) and \(q_{2}=-q .\) Express your answers in terms of \(q\)

The electric potential energy stored in the capacitor of a defibrillator is \(73 \mathrm{~J}\), and the capacitance is \(120 \mu \mathrm{F}\). What is the potential difference that exists across the capacitor plates?

An electron is released at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see the drawing), (a) As the electron gains kinetic energy, does its electric potential energy increase or decrease? Why? (b) The difference in the electron's electric potential energy between the positive and negative plates is EPE positive \(-\mathrm{EPE}_{\text {negative. }}\) How is this difference related to the charge on the electron \((-e)\) and to the difference \(V\) positive \(-V_{\text {negative in the electric potential between }}\) the plates? (c) How is the potential difference \(V\) positive \(-V_{\text {negative }}\) related to the electric field within the capacitor and the displacement of the positive plate relative to the negative plate? The plates of a parallel plate capacitor are separated by a distance of \(1.2 \mathrm{~cm}\), and the electric field within the capacitor has a magnitude of \(2.1 \times 10^{6} \mathrm{~V} / \mathrm{m}\). An electron starts from rest at the negative plate and accelerates to the positive plate. What is the kinetic energy of the electron just as the electron reaches the positive plate?
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