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Chapter 18: Problem 26

\(\mathrm{A}+1.50 \mu \mathrm{C}\) point charge is sitting at the origin. (a) What is the radial distance between the 500 V equipotential surface and the \(1000 \mathrm{~V}\) surface? (b) What is the distance between the \(1000 \mathrm{~V}\) surface and the \(1500 \mathrm{~V}\) surface? (c) Explain why the answers to (a) and (b) are not the same.

Short Answer

Expert verified
(a) Compute \(r_2 - r_1\); (b) Compute \(r_1 - r_3\); (c) Distances differ due to diminishing field strength with distance.

Step by step solution

01

Understanding the Equipotential Surfaces

Equipotential surfaces are surfaces where the electric potential is constant. For a point charge, the equipotential surfaces are spheres centered on the charge. The electric potential due to a point charge at distance \( r \) is given by \( V = \frac{kQ}{r} \), where \( k \) is the electrostatic constant \( 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \) and \( Q \) is the charge.
02

Set Up Equation for Radial Distances

We have two equations related to the equipotential surfaces due to the charge \(+1.50 \mu \text{C}\), which need to be solved. For \(1000 \text{ V}\) equipotential surface: \( 1000 = \frac{(8.99 \times 10^9)(1.50 \times 10^{-6})}{r_1} \). For \(500 \text{ V}\) equipotential surface: \( 500 = \frac{(8.99 \times 10^9)(1.50 \times 10^{-6})}{r_2} \).
03

Calculate Radial Distance for 1000 V Surface

Rearrange the equation \( 1000 = \frac{(8.99 \times 10^9)(1.50 \times 10^{-6})}{r_1} \) to solve for \( r_1 \): \( r_1 = \frac{(8.99 \times 10^9)(1.50 \times 10^{-6})}{1000} \). Calculate to find \( r_1 \).
04

Calculate Radial Distance for 500 V Surface

Rearrange the equation \( 500 = \frac{(8.99 \times 10^9)(1.50 \times 10^{-6})}{r_2} \) to solve for \( r_2 \): \( r_2 = \frac{(8.99 \times 10^9)(1.50 \times 10^{-6})}{500} \). Calculate to find \( r_2 \).
05

Compute Distance Between 500 V and 1000 V Surfaces

The radial distance between two equipotential surfaces is the difference between their radial distances. Therefore, distance (a) is \( r_2 - r_1 \).
06

Calculate Radial Distance for 1500 V Surface

Use the formula \( 1500 = \frac{(8.99 \times 10^9)(1.50 \times 10^{-6})}{r_3} \) to find \( r_3 \); rearrange to \( r_3 = \frac{(8.99 \times 10^9)(1.50 \times 10^{-6})}{1500} \). Calculate \( r_3 \).
07

Compute Distance Between 1000 V and 1500 V Surfaces

The distance (b) is the difference between the radial distances \( r_1 \) and \( r_3 \); \( r_1 - r_3 \).
08

Explaining Different Distances between Equipotential Surfaces

The distances differ because as you get farther from the charge, the rate at which potential changes with respect to distance decreases due to the inverse relationship \( V \propto \frac{1}{r} \), leading to a decrease in electric field strength.

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Key Concepts

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

Electric Potential
Electric potential is a key concept when studying electric fields and charges. It represents the amount of work needed to move a unit positive charge from a reference point (generally infinity) to a specific point in an electric field without any acceleration.
For a point charge, the electric potential at a distance \( r \) from the charge is given by:
  • \( V = \frac{kQ}{r} \)
where:
  • \( V \) is the electric potential measured in volts (V),
  • \( k \) is the electrostatic constant, approximately \( 8.99 \times 10^9 \) NmyC虏,
  • \( Q \) is the charge (in this exercise, \(1.50 \mu \text{C} \)),
  • \( r \) is the radial distance from the charge to the point of interest.
The concept of electric potential is crucial for understanding how energy is transferred in electric fields.
It helps visualize energy levels similar to contours on a topographical map, where higher potentials correspond to higher energy levels.
Point Charge
A point charge is a hypothetical charge that is considered to be located at a single point in space. This idealization simplifies the analysis of electric fields and electric potential. In practical terms, a point charge is a charge located at a small region such that compared to other relevant distances, it can be considered as a point. Key characteristics of a point charge include:
  • The electric field it produces radiates outward spherically from the point of the charge.
  • The strength of the electric field depends on the distance from the charge.
  • Its potential field is determined solely by its size and location, according to the equation \( V = \frac{kQ}{r} \).
Understanding point charges is pivotal in physics as many calculations regarding real-world charge distributions can be approximated or simplified using the concept of point charges.
Electrostatic Constant
The electrostatic constant, also known as Coulomb's constant, is a fundamental constant that quantifies the magnitude of electrostatic force between two point charges in vacuum. It is denoted by \( k \) and its approximate value is \( 8.99 \times 10^9 \text{ Nm}^2/ ext{C}^2 \). This constant is key in Coulomb's Law, which is expressed as:
  • \( F = \frac{k |Q_1 Q_2|}{r^2} \)
where:
  • \( F \) is the electrostatic force between two charges,
  • \( |Q_1 Q_2| \) is the product of the magnitudes of the two point charges,
  • \( r \) is the distance separating the charges.
The value of the electrostatic constant ensures that the computed forces between charges are sensible when using standard units. The constant is pivotal for translating electric potential and fields into measurable, real-world effects.
Radial Distance
In the context of electric fields, radial distance refers to how far a particular point is from the source charge, here a point charge. In mathematical terms, it is the straight-line distance measured from the charge to any given point of interest. It is crucial for understanding equipotential surfaces, which are imaginary spheres around a point charge at which the electric potential is equal. Key points about radial distance include:
  • The electric field strength and potential diminish with increasing \( r \), as \( V \) is inversely related to \( r \) by the formula \( V = \frac{kQ}{r} \).
  • Equipotential surfaces allow us to visualize how potential drops with increasing radius from the charge.
  • The radial distance between two equipotential surfaces indicates how "spread out" the potential change is over space.
Hence, radial distance is an essential factor when calculating the exact electric potential at specific points around a point charge.

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Most popular questions from this chapter

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. \(22 e\)

A parallel-plate air capacitor has a capacitance of \(920 \mathrm{pF}\). The charge on each plate is \(2.55 \mu \mathrm{C}\). (a) What is the potential difference between the plates? (b) If the charge is kept constant, what will be the potential difference between the plates if the separation is doubled? (c) How much work is required to double the separation?

When two point charges are a distance \(R\) apart, their potential energy is \(-2.0 \mathrm{~J}\). How far (in terms of \(R\) ) should they be from each other so that their potential energy is \(-6.0 \mathrm{~J} ?\)

A \(10.0 \mu \mathrm{F}\) parallel-plate capacitor with circular plates is connected to a \(12.0 \mathrm{~V}\) battery. (a) What is the charge on each plate? (b) How much charge would be on the plates if their separation were doubled while the capacitor remained connected to the battery? (c) How much charge would be on the plates if the capacitor were connected to the \(12.0 \mathrm{~V}\) battery after the radius of each plate was doubled without changing their separation?

A parallel-plate capacitor having plates \(6.0 \mathrm{~cm}\) apart is connected across the terminals of a \(12 \mathrm{~V}\) battery. (a) Being as quantitative as you can, describe the location and shape of the equipotential surface that is at a potential of \(+6.0 \mathrm{~V}\) relative to the potential of the negative plate. Avoid the edges of the plates. (b) Do the same for the equipotential surface that is at \(+2.0 \mathrm{~V}\) relative to the negative plate. (c) What is the potential gradient between the plates?
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