/*! 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] | 91Ó°ÊÓ

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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 \(\mathrm{V}\) and (b) 30.0 \(\mathrm{V}\) ? Take the potential to be zero at an infinite distance from the charge.

Short Answer

Expert verified
Distances are 0.25 m for 90 V and 0.75 m for 30 V.

Step by step solution

01

Understand the Formula for Electric Potential

The electric potential \( V \) due to a point charge \( q \) at a distance \( r \) is given by the formula:\[V = \frac{k \cdot q}{r}\]where \( k \) is Coulomb's constant, approximately \(8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2\). The problem tells us to find \( r \) for particular values of \( V \).
02

Solve for Distance When Potential is 90.0 V

Rearrange the formula to solve for \( r \):\[r = \frac{k \cdot q}{V}\]Substitute the given values, \( q = 2.50 \times 10^{-11} \, \text{C} \) and \( V = 90.0 \, \text{V} \):\[r = \frac{8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \times 2.50 \times 10^{-11} \, \text{C}}{90.0 \, \text{V}}\]Calculate \( r \):\[r \approx 0.25 \, \text{m}\]
03

Solve for Distance When Potential is 30.0 V

Use the same rearranged formula to find \( r \) for \( V = 30.0 \, \text{V} \):\[r = \frac{8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \times 2.50 \times 10^{-11} \, \text{C}}{30.0 \, \text{V}}\]Calculate \( r \):\[r \approx 0.75 \, \text{m}\]
04

Verify and Conclude

Both distances have been calculated under the assumption that potential decreases with increasing distance, aligning with the physical expectation for an isolated point charge. The calculations are consistent and adhere to the physical principles governing electric potential.

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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 theoretical concept in physics indicating an electric charge concentrated at a single point in space. Although it doesn't actually exist as a physical entity, it serves as a useful model for understanding electric forces and potential in simpler terms. By thinking of electric charges in this way, complex systems can be broken down and analyzed more easily.

Understanding point charges helps in predicting how electric fields behave around charged objects. Since the charge is concentrated at a specific location, the effects of that charge, like the electric field or potential, can be calculated in a straightforward manner using mathematical formulas. This approach provides a foundation for more complex concepts in electromagnetism.
Coulomb's Law
Coulomb's Law is central to the study of electric charges and their interactions. It states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

The mathematical expression for Coulomb's Law is:
  • Formula: \( F = k \frac{|q_1 q_2|}{r^2} \)
  • Where:
    • \( F \) is the force between the charges,
    • \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
    • \( r \) is the distance between the charges,
    • \( k \) is Coulomb's constant.
In the context of electric potential, Coulomb's Law helps determine how much work would be required to bring a charge from infinity to a point in space due to another charge. This is key to understanding electric potential, as it directly relates to the amount of energy per unit charge at a point.
Electric Field
The electric field is a vector field that represents the force a charge would experience at any point in space. It gives a complete picture of how a point charge influences its surrounding environment and defines how other charges will react when placed in an area.

  • The formula for the electric field \( E \) due to a point charge \( q \) is:\[ E = \frac{k \, q}{r^2} \]
  • This shows that the field strength decreases with the square of the distance \( r \) from the charge causing it.
Understanding electric fields is crucial as they describe how charges interact over a distance without physical contact. The strength and direction of the field give insight into potential energy changes and forces applied to other charges within the field's domain.
Physics Education
Physics education focuses on understanding the fundamental principles that govern the natural world. Concepts like electric potential, point charges, and Coulomb's Law form the backbone of classical physics study and pave the way for exploring advanced topics.

  • Effective learning strategies include:
    • Breaking down complex theories into simpler components.
    • Applying real-world problems to theoretical concepts.
    • Engaging in hands-on experiments to see theory in action.
  • Thinking critically about results and their implications encourages deeper comprehension and retention of material.
By mastering these basic concepts, students gain a solid foundation to build upon as they advance in their study of physics and related sciences. The structured step-by-step methods encourage students to connect the dots between different areas, fostering a cohesive understanding of the physical universe.

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Most popular questions from this chapter

A very long insulating cylinder of charge of radius 2.50 \(\mathrm{cm}\) carries a uniform linear density of 15.0 \(\mathrm{nC} / \mathrm{m}\) . If you put one probe of a voltmeter at the surface, how far from the surface must the other probe be placed so that the voltmeter reads 175 \(\mathrm{V} ?\)

At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are 4.98 \(\mathrm{V}\) and \(12.0 \mathrm{V} / \mathrm{m},\) respectively. (Take the potential to be zero at infinity.) (a) What is the distance to the point charge? (b) What is the magnitude of the charge? (c) Is the electric field directed toward or away from the point charge?

An infinitely long line of charge has linear charge density \(5.00 \times 10^{-12} \mathrm{C} / \mathrm{m} .\) A proton (mass \(1.67 \times 10^{-27} \mathrm{kg}\) . charge \(+1.60 \times 10^{-19} \mathrm{C} )\) is 18.0 \(\mathrm{cm}\) from the line and moving directly toward the line at \(1.50 \times 10^{3} \mathrm{m} / \mathrm{s}\) (a) Calculate the proton's initial kinetic energy. (b) How close does the proton get to the line of charge? (Hints See Example \(23.10 . )\)

A very long wire carries a uniform linear charge density \(\lambda\) . Using a voltmeter to measure potential difference, you find that when one probe of the meter is placed 2.50 \(\mathrm{cm}\) from the wire and the other probe is 1.00 \(\mathrm{cm}\) farther from the wire, the meter reads 575 \(\mathrm{V}\) . (a) What is \(\lambda 2\) (b) If you now place one probe at 3.50 \(\mathrm{cm}\) from the wire and the other probe 1.00 \(\mathrm{cm}\) farther away, will the voltmeter read 575 \(\mathrm{V} ?\) If not, will it read more or less than 575 \(\mathrm{V}\) ? Why? (c) If you place both probes 3.50 \(\mathrm{cm}\) from the wire but 17.0 \(\mathrm{cm}\) from each other, what will the voltmeter read?

A positive charge \(q\) is fixed at the point \(x=0, y=0,\) and a negative charge \(-2 q\) is fixed at the point \(x=a, y=0\) . (a) Show the positions of the charges in a diagram. (b) Derive an expression for the potential \(V\) at points on the \(x\) -axis as a function of the coordinate \(x\) . Take \(V\) to be zero at an infinite distance from the charges. (c) At which positions on the \(x\) -axis is \(V=0 ?(\text { d) Graph }\) \(V\) at points on the \(x\) -axis as a function of \(x\) in the range from \(x=-2 a\) to \(x=+2 a\) (e) What does the answer to part \((b)\) become when \(x \gg a ?\) Explain why this result is obtained.

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