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Chapter 23: Problem 18

An electron moves away from a proton. Describe how the potential it encounters changes. Describe how its potential energy is changing.

Short Answer

Expert verified
Answer: As the electron moves away from the proton, the electric potential decreases and the electric potential energy of the system also decreases.

Step by step solution

01

Explain the formula for electric potential

The electric potential between two charges is given by the formula: V = k * Q / r where V is the electric potential, k is the electrostatic constant (approximately 8.99 * 10^9 N m^2 C^-2), Q is the charge (in this case, the charge of a proton), and r is the distance between the charges.
02

Describe how the electric potential changes when the electron moves away from the proton

As the distance r between the electron and the proton increases, the electric potential V will decrease according to the formula V = k * Q / r. This means that the electron encounters a lower potential as it moves away from the proton.
03

Explain the formula for electric potential energy

The electric potential energy between two charges is given by the formula: U = k * Q1 * Q2 / r where U is the electric potential energy, Q1 and Q2 are the charges (in this case, the charges of the electron and proton), and r is the distance between the charges.
04

Describe how the electric potential energy changes when the electron moves away from the proton

As the distance r between the electron and the proton increases, their electric potential energy U will decrease according to the formula U = k * Q1 * Q2 / r. This means that the potential energy of the system is lower as the electron moves away from the proton.

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Most popular questions from this chapter

Two parallel plates are held at potentials of \(+200.0 \mathrm{~V}\) and \(-100.0 \mathrm{~V}\). The plates are separated by \(1.00 \mathrm{~cm}\). a) Find the electric field between the plates. b) An electron is initially placed halfway between the plates. Find its kinetic energy when it hits the positive plate.

Using Gauss's Law and the relation between electric potential and electric field, show that the potential outside a uniformly charged sphere is identical to the potential of a point charge placed at the center of the sphere and equal to the total charge of the sphere. What is the potential at the surface of the sphere? How does the potential change if the charge distribution is not uniform but has spherical (radial) symmetry?

In molecules of gaseous sodium chloride, the chloride ion has one more electron than proton, and the sodium ion has one more proton than electron. These ions are separated by about \(0.24 \mathrm{nm}\). How much work would be required to increase the distance between these ions to \(1.0 \mathrm{~cm} ?\)

A charge of \(0.681 \mathrm{nC}\) is placed at \(x=0 .\) Another charge of \(0.167 \mathrm{nC}\) is placed at \(x_{1}=10.9 \mathrm{~cm}\) on the \(x\) -axis. a) What is the combined electrostatic potential of these two charges at \(x=20.1 \mathrm{~cm},\) also on the \(x\) -axis? b) At which point(s) on the \(x\) -axis does this potential have a minimum?

The electric potential energy of a continuous charge distribution can be found in a way similar to that used for systems of point charges in Section \(23.6,\) by breaking the distribution up into suitable pieces. Find the electric potential energy of an arbitrary spherically symmetrical charge distribution, \(\rho(r) .\) Do not assume that \(\rho(r)\) represents a point charge, that it is constant, that it is piecewise-constant, or that it does or does not end at any finite radius, \(r\). Your expression must cover all possibilities. Your expression may include an integral or integrals that cannot be evaluated without knowing the specific form of \(\rho(r) .\) (Hint: A spherical pearl is built up of thin layers of nacre added one by one.)
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