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Chapter 18: Problem 5

Two stationary point charges of \(+3.00 \mathrm{nC}\) and \(+2.00 \mathrm{nC}\) are separated by a distance of \(50.0 \mathrm{~cm} .\) An electron is released from rest at a point midway between the charges and moves along the line connecting them. What is the electric potential energy for the electron when it is (a) at the midpoint and (b) \(10.0 \mathrm{~cm}\) from the \(+3.00 \mathrm{nC}\) charge?

Short Answer

Expert verified
(a) \(-2.88 \times 10^{-17} \text{ J}\), (b) \(-5.03 \times 10^{-17} \text{ J}\).

Step by step solution

01

Understanding the Problem

We have two charges, \(+3.00\, \text{nC}\) and \(+2.00\, \text{nC}\) that are separated by \(50 \, \text{cm}\). An electron is released at the midpoint. We need to compute the electric potential energy at two points on the line connecting the charges.
02

Calculating the potential energy at the midpoint

Since the electron is at the midpoint between the two charges, the distance from each charge to the midpoint is \(25.0\, \text{cm} = 0.25\, \text{m}\). The electric potential \(V\) due to a point charge \(q\) at a distance \(r\) is given by: \(V = \frac{k \cdot q}{r}\), where \(k = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2\). The total potential \(V_{\text{total}}\) at midpoint due to both charges is: \[ V_{\text{total}} = V_1 + V_2 = \frac{k \cdot 3.00 \times 10^{-9} \text{ C}}{0.25} + \frac{k \cdot 2.00 \times 10^{-9} \text{ C}}{0.25} \] The electric potential energy \(U\) for the electron is \(U = -e \cdot V_{\text{total}}\), where \(e = 1.60 \times 10^{-19} \text{ C}\).
03

Potential energy calculation at 10 cm from the +3.00 nC charge

At \(10.0\, \text{cm}\) from the \(+3.00\, \text{nC}\) charge, the electron is \(40.0\, \text{cm} = 0.40\, \text{m}\) from the \(+2.00\, \text{nC}\) charge. Calculate the potential at this new point:\[ V_{\text{new}} = \frac{k \cdot 3.00 \times 10^{-9}}{0.10} + \frac{k \cdot 2.00 \times 10^{-9}}{0.40} \] Again, the electric potential energy \(U\) here is given by \(U = -e \cdot V_{\text{new}}\).
04

Numerical Computation

Use the value of \(k\) and perform calculations:**Midpoint:**\[ V_{\text{total}} = \frac{8.99 \times 10^9 \cdot 3.00 \times 10^{-9}}{0.25} + \frac{8.99 \times 10^9 \cdot 2.00 \times 10^{-9}}{0.25} = 107.88 + 71.92 = 179.8 \, \text{V}\]\[ U = - (1.60 \times 10^{-19}) \times 179.8 = -2.8768 \times 10^{-17} \, \text{J}\]**10 cm from \(+3.00 \, \text{nC}\):**\[ V_{\text{new}} = \frac{8.99 \times 10^9 \cdot 3.00 \times 10^{-9}}{0.10} + \frac{8.99 \times 10^9 \cdot 2.00 \times 10^{-9}}{0.40} = 269.7 + 44.95 = 314.65 \, \text{V}\]\[ U = - (1.60 \times 10^{-19}) \times 314.65 = -5.0344 \times 10^{-17} \, \text{J}\]

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Key Concepts

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

Understanding Point Charges
Point charges are electric charges that are assumed to exist at a single point in space. This concept is essential for calculating the behavior of electric fields and potentials because it simplifies the mathematical modeling of the interactions between charges.
  • Point charges can be positive or negative.
  • The interaction between multiple point charges can be calculated using the principles of superposition.
  • In the original exercise, the point charges are stationary, meaning they do not move or change position.
Understanding how point charges interact helps to lay the foundation for calculating other properties like electric potential and force between charges.
Exploring Electric Potential
Electric potential, often denoted as \(V\), is a measure of the potential energy per unit charge at a given point in an electric field. It is important because it tells us how much work must be done to move a charge within that field.
  • The potential due to a point charge is calculated as \(V = \frac{k \cdot q}{r}\), where \(k\) is Coulomb's constant, \(q\) is the charge, and \(r\) is the distance from the charge.
  • Electric potential is a scalar quantity, meaning it only has magnitude and no direction.
  • In the exercise, the total electric potential midway between two charges is obtained by adding the potentials due to individual charges, illustrating the superposition principle.
Identifying the electric potential at various points allows us to compute the electric potential energy, especially concerning electron motion.
Electron Motion in Electric Fields
Electrons are negatively charged particles, and they naturally move towards regions of higher electric potential. This movement is influenced by the electric forces exerted by point charges.
  • When released in an electric field, electrons move according to the forces exerted by surrounding charges.
  • The electric potential energy for an electron \(U\) is computed as \(U = -e \cdot V\), where \(e\) is the electron's charge.
  • The negative sign indicates that as the potential \(V\) increases, the electron's potential energy becomes more negative.
Understanding how electrons move in an electric field helps us predict their path and determine their potential energy at different positions, as shown in the original solution.
Coulomb's Law and Force Between Charges
Coulomb's Law is a fundamental principle that describes the force between two point charges. It鈥檚 important for understanding not only how charges interact with each other but also how they influence electric potential and energy.
  • The force between two charges is given by \( F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \), where \(q_1\) and \(q_2\) are the magnitudes of the charges, \(r\) is the distance between them, and \(k\) is Coulomb's constant.
  • This force is attractive if the charges are of opposite signs and repulsive if they are of the same sign.
  • Coulomb's Law helps explain why the electron moves towards the region of higher potential energy in our original exercise.
By using Coulomb's Law, we can better understand the dynamics of charged particles like electrons in the context of electric fields and forces.

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Most popular questions from this chapter

Electrical sensitivity of sharks. Certain sharks can detect an electric field as weak as \(1.0 \mu \mathrm{V} / \mathrm{m}\). To grasp how weak this field is, if you wanted to produce it between two parallel metal plates by connecting an ordinary \(1.5 \mathrm{~V}\) AA battery across these plates, how far apart would the plates have to be?

A point charge \(Q=+4.60 \mu \mathrm{C}\) is held fixed at the origin. A second point charge \(q=+1.20 \mu \mathrm{C}\) with mass of \(2.80 \times 10^{-4} \mathrm{~kg}\) is placed on the \(x\) axis, \(0.250 \mathrm{~m}\) from the origin. (a) What is the electric potential energy \(U\) of the pair of charges? (Take \(U\) to be zero when the charges have infinite separation.) (b) The second point charge is released from rest. What is its speed when its distance from the origin is (i) \(0.500 \mathrm{~m} ;\) (ii) \(5.00 \mathrm{~m} ;\) (iii) \(50.0 \mathrm{~m} ?\)

Cell membranes. Cell membranes (the walled enclosure around a cell) are typically about \(7.5 \mathrm{nm}\) thick. They are partially permeable to allow charged material to pass in and out, as needed. Equal but opposite charge densities build up on the inside and outside faces of such a membrane, and these charges prevent additional charges from passing through the cell wall. We can model a cell membrane as a parallel-plate capacitor, with the membrane itself containing proteins embedded in an organic material to give the membrane a dielectric constant of about 10\. (See Figure \(18.47 .)\) (a) What is the capacitance per square centimeter of such a cell wall? (b) In its normal resting state, a cell has a potential difference of \(85 \mathrm{mV}\) across its membrane. What is the electric field inside this membrane?

A parallel-plate capacitor is to be constructed by using, as a dielectric, rubber with a dielectric constant of 3.20 and a dielectric strength of \(20.0 \mathrm{MV} / \mathrm{m}\). The capacitor is to have a capacitance of \(1.50 \mathrm{nF}\) and must be able to withstand a maximum potential difference of \(4.00 \mathrm{kV}\). What is the minimum area the plates of this capacitor can have?

A helium ion \(\left(\mathrm{He}^{++}\right)\) that comes within about \(10 \mathrm{fm}\) of the center of the nucleus of an atom in the sample may induce a nuclear reaction instead of simply scattering. Imagine a helium ion with a kinetic energy of \(3.0 \mathrm{MeV}\) heading straight toward an atom at rest in the sample. Assume that the atom stays fixed. What minimum charge can the nucleus of the atom have such that the helium ion gets no closer than \(10 \mathrm{fm}\) from the center of the atomic nucleus? (1 \(\mathrm{fm}=1 \times 10^{-15} \mathrm{~m},\) and \(e\) is the magnitude of the charge of an electron or a proton. A. \(2 e\) B. \(11 e\) C. \(20 e\) D. \(22 e\)
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