/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 A point charge has a charge of \... [FREE SOLUTION] | 91影视

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Chapter 18: Problem 19

A point charge has a charge of \(2.50 \times 10^{-11} \mathrm{C}\). At what distance from the point charge is the electric potential (a) \(90.0 \mathrm{~V} ?\) (b) \(30.0 \mathrm{~V}\) ? Take the potential to be zero at an infinite distance from the charge.

Short Answer

Expert verified
(a) Approx. 0.250 m; (b) Approx. 0.749 m.

Step by step solution

01

Understanding the Problem

We have a point charge with a given charge value and want to find the distance at which the electric potential is specified values. The point charge has a charge of \( q = 2.50 \times 10^{-11} \mathrm{C} \). We need to find the distance \( r \) at which the electric potential \( V \) is either \( 90.0 \mathrm{~V} \) or \( 30.0 \mathrm{~V} \).
02

Electric Potential Formula

The electric potential \( V \) at a distance \( r \) from a point charge \( q \) is given by the formula:\[ V = \frac{k \cdot q}{r} \]where \( k \) is the electrostatic constant \( k \approx 8.99 \times 10^9 \mathrm{~N}\cdot\mathrm{m}^2/\mathrm{C}^2 \).
03

Rearrange the Formula

To find the distance \( r \), rearrange the formula to solve for \( r \):\[ r = \frac{k \cdot q}{V} \]
04

Calculate Distance for \( V = 90.0 \mathrm{~V} \)

Substitute the given values into the rearranged formula for \( V = 90.0 \mathrm{~V} \):\[ r = \frac{8.99 \times 10^9 \mathrm{~N}\cdot\mathrm{m}^2/\mathrm{C}^2 \times 2.50 \times 10^{-11} \mathrm{C}}{90.0 \mathrm{~V}} \]Calculate to find \( r \).
05

Calculate Distance for \( V = 30.0 \mathrm{~V} \)

Now, substitute the given values into the rearranged formula for \( V = 30.0 \mathrm{~V} \):\[ r = \frac{8.99 \times 10^9 \mathrm{~N}\cdot\mathrm{m}^2/\mathrm{C}^2 \times 2.50 \times 10^{-11} \mathrm{C}}{30.0 \mathrm{~V}} \]Calculate to find \( r \).
06

Compute the Results

For \( V = 90.0 \mathrm{~V} \):\[ r = \frac{8.99 \times 10^9 \cdot 2.50 \times 10^{-11}}{90.0} \approx 0.2497 \mathrm{~m} \approx 0.250 \mathrm{~m} \]For \( V = 30.0 \mathrm{~V} \):\[ r = \frac{8.99 \times 10^9 \cdot 2.50 \times 10^{-11}}{30.0} \approx 0.7491 \mathrm{~m} \approx 0.749 \mathrm{~m} \]

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Key Concepts

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

Point Charge
A point charge is a simple model used to represent a charged particle. Consider it as a tiny particle with a certain amount of electric charge. In physics, charges are usually measured in Coulombs (C). This model is incredibly helpful for simplifying problems because it allows us to focus on the principles of electrostatics without getting tangled up in complexities like the shape and distribution of the charge.
  • In real-life scenarios, point charges are just an idealization, but they are useful for understanding fundamental concepts.
  • The electric potential around a point charge decreases as you move further away from it.
Understanding how a point charge works is crucial for exploring the effects of electric fields and potentials in various physics problems.
Coulomb's Law
Coulomb's Law is a method used to calculate the electrostatic force between two point charges. However, it also gives us insights into how charges interact over a distance. It states that the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
  • Formula for electric force: \( F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \) where \( k \) is the electrostatic constant.
  • According to Coulomb's Law, like charges repel, while opposite charges attract.
By using this law, one can calculate not just force but also delve deeper into understanding electric fields surrounding point charges and their interactions.
Electrostatic Constant
The electrostatic constant, often denoted as \( k \) in equations, is a fundamental constant in physics that plays a major role in Coulomb's Law. It's also referred to as Coulomb's constant. This constant helps relate the forces between charges to their distances and magnitudes. The value of the electrostatic constant is approximately \( 8.99 \times 10^9 \; \text{N} \cdot \text{m}^2/\text{C}^2 \).
  • This constant is crucial for calculations involving electric forces and potentials.
  • Understanding \( k \) is important for solving problems related to electric fields and potentials because it provides a standard coefficient used in equations.
The consistency of the electrostatic constant across environments simplifies calculations when dealing with electric potentials and forces.
Distance Calculation
Distance calculation in electrostatics often involves using the rearranged formula derived from the electric potential equation. For a point charge, electric potential \( V \) is calculated as: \[ V = \frac{k \cdot q}{r} \] where \( q \) is the charge, and \( r \) is the distance from the charge. By rearranging this equation, you can solve for \( r \) in terms of known quantities. This leads to: \( r = \frac{k \cdot q}{V} \).
  • Calculate the distance \( r \) given a specific potential \( V \) using known values of \( q \) and \( k \).
  • In exercises, it's common to be given values for the potential and charge while solving for the variable \( r \).
This understanding is key when determining how potential varies with distance, providing insights into how electric fields operate at different distances from a charge.

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Most popular questions from this chapter

Dipole. A dipole is located on a sheet of paper. (a) In the plane of that paper, carefully sketch the electric field lines for this dipole. (b) Use your field lines in part (a) to sketch the equipotential curves where the equipotential surfaces intersect the paper.

Electric eels. Electric eels and electric fish generate large potential differences that are used to stun 9 enemies and prey. These potentials are produced by cells that each can generate 0.10 V. We can plausibly model such cells as charged capacitors. (a) How should these cells be connected \((\mathrm{in}\) series or in parallel) to produce a total potential of more than 0.10 \(\mathrm{V} ?\) (b) Using the connection in part (a), how many cells must be connected together to produce the 500 \(\mathrm{V}\) surge of the electric eel?

How much charge does a 12 \(\mathrm{V}\) battery have to supply to fully charge a 2.5\(\mu \mathrm{F}\) capacitor and a 5.0\(\mu \mathrm{F}\) capacitor when they're (a) in parallel, (b) in series? (c) How much energy does the battery have to supply in each case?

You make a capacitor by cutting the \(15.0-\mathrm{cm}\) -diameter bottoms out of two aluminum pie plates, separating them by 3.50 \(\mathrm{mm},\) and connecting them across a \(6.00-\mathrm{V}\) battery. (a) What's the capacitance of your capacitor? (b) If you disconnect the battery and separate the plates to a distance of \(3.50 \mathrm{~cm}\) without discharging them, what will be the potential difference between them?

A parallel-plate air capacitor has a capacitance of 920 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?
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