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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

A solid metal ball with a radius of \(3.00 \mathrm{~m}\) has a charge of \(4.00 \mathrm{mC}\). If the electric potential is zero far away from the ball, what is the electric potential at each of the following positions? a) at \(r=0 \mathrm{~m},\) the center of the ball b) at \(r=3.00 \mathrm{~m},\) on the surface of the ball c) at \(r=5.00 \mathrm{~m}\)

Derive an expression for electric potential along the axis (the \(x\) -axis) of a disk with a hole in the center, as shown in the figure, where \(R_{1}\) and \(R_{2}\) are the inner and outer radii of the disk. What would the potential be if \(R_{1}=0 ?\)

Two metal spheres of radii \(r_{1}=10.0 \mathrm{~cm}\) and \(r_{2}=\) \(20.0 \mathrm{~cm},\) respectively, have been positively charged so that both have a total charge of \(100, \mu C\) a) What is the ratio of their surface charge distributions? b) If the two spheres are connected by a copper wire, how much charge flows through the wire before the system reaches equilibrium?

A charge \(Q=\) \(+5.60 \mu C\) is uniformly distributed on a thin cylindrical plastic shell. The radius, \(R\), of the shell is \(4.50 \mathrm{~cm}\). Calculate the electric potential at the origin of the \(x y\) -coordinate system shown in the figure. Assume that the electric potential is zero at points infinitely far away from the origin.

The electric field, \(\vec{E}(\vec{r}),\) and the electric potential \(V(\vec{r}),\) are calculated from the charge distribution, \(\rho(\vec{r}),\) by integrating Coulomb's Law and then the electric field. In the other direction, the field and the charge distribution are determined from the potential by suitably differentiating. Suppose the electric potential in a large region of space is given by \(V(r)=V_{0} \exp \left(-r^{2} / a^{2}\right),\) where \(V_{0}\) and \(a\) are constants and \(r=\sqrt{x^{2}+y^{2}+z^{2}}\) is the distance from the origin. a) Find the electric field \(\vec{E}(\vec{r})\) in this region. b) Determine the charge density \(\rho(\vec{r})\) in this region, which gives rise to the potential and field. c) Find the total charge in this region. d) Roughly sketch the charge distribution that could give rise to such an electric field.
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