/*! 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 17 At what separation is the electr... [FREE SOLUTION] | 91Ó°ÊÓ

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At what separation is the electrostatic force between a \(+11.2 \cdot \mu C\) point charge and \(a+29.1-\mu C\) point charge equal in magnitude to 1.57 N?

Short Answer

Expert verified
The separation distance is approximately 0.1369 meters.

Step by step solution

01

Understand the Given Problem

We are given two point charges, one with a charge of \(+11.2 \ \mu C\) and the other with \(+29.1 \ \mu C\). We need to find the separation distance at which the electrostatic force between them equals 1.57 N.
02

Convert Charge Units

Convert the microcoulomb (\(\mu C\)) units to coulombs (C) for both charges. \[11.2 \ \mu C = 11.2 \times 10^{-6} \, C\]\[29.1 \ \mu C = 29.1 \times 10^{-6} \, C\]
03

Use Coulomb's Law Formula

Coulomb’s Law states that the electrostatic force \(F\) between two point charges is given by the formula: \[F = \frac{k \, |q_1 q_2|}{r^2}\]where \(k\) is Coulomb's constant, \(9 \times 10^9 \, N \, m^2/C^2\), \(q_1\) and \(q_2\) are the charges, and \(r\) is the separation distance.
04

Plug Values into Coulomb's Law

Substitute the absolute values of the charges, the given force, and Coulomb's constant into the formula: \[1.57 = \frac{9 \times 10^9 \, \cdot \, (11.2 \times 10^{-6} \, C) \cdot (29.1 \times 10^{-6} \, C)}{r^2}\]
05

Solve for r

Rearrange the formula to solve for \(r^2\): \[r^2 = \frac{9 \times 10^9 \, \cdot \, (11.2 \times 10^{-6} \cdot 29.1 \times 10^{-6})}{1.57}\]Calculate the numerator: \[9 \times 10^9 \, \cdot \, 11.2 \times 10^{-6} \cdot 29.1 \times 10^{-6} = 2.943048 \times 10^{-2}\]So,\[r^2 = \frac{2.943048 \times 10^{-2}}{1.57} \approx 0.01874479\]Now take the square root to find \(r\): \[r \approx \sqrt{0.01874479} \approx 0.1369 \ m\]
06

Conclusion

The separation distance \(r\) must be approximately 0.1369 meters for the force to be 1.57 N between the given point charges.

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Key Concepts

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

Electrostatic Force
Electrostatic force is a fundamental interaction between charged objects. This force acts between particles that have an electric charge. One of the essential characteristics of this force is that, unlike gravitational force which is always attractive, electrostatic force can either attract or repel the charges involved.

The magnitude of the electrostatic force is determined by Coulomb's Law. The law states that the force between two point charges is directly proportional to the product of the absolute values of the charges and inversely proportional to the square of the separation distance between them. The formula is given by:
  • Coulomb's Law formula: \[F = \frac{k \times |q_1 q_2|}{r^2}\] where \(k\) is the Coulomb's constant \((9 \times 10^9 \, N \, m^2/C^2)\).
This means that as the distance between the charges increases, the force decreases rapidly.Understanding electrostatic force is crucial because it plays a significant role in various phenomena and applications, from simple electrostatic experiments to complex electronic systems.
Point Charges
Point charges are idealized charges often used in physics to model the position and effect of a charged particle or object. The model helps simplify complex real-world problems by considering the charge as being concentrated at a single point in space.

Some of the important characteristics of point charges include:
  • They are dimensionless, meaning the size of the charge does not affect the calculation of forces.
  • Point charges can be positive or negative, determining how they interact with other charges (attract or repel).
  • They significantly simplify the application of Coulomb's Law, focusing just on charge values and separation distance.
In exercises involving point charges, like the one provided, it is essential to use these characteristics to focus purely on the interaction determined by their charge values and the distance between them. Understanding point charges allows students to practice calculating forces without the complexities introduced by the real shapes and sizes of charged objects.
Separation Distance
Separation distance, often denoted as \(r\) in formulas such as Coulomb's Law, is the straight-line distance between two charges. It is a crucial factor in determining the electrostatic force exerted on or by the charges.

Some key points about separation distance include:
  • According to Coulomb’s Law, the force is inversely proportional to the square of this distance. Hence, even a small change in distance can significantly impact the force.
  • To find the exact separation distance for a specific force, rearrange Coulomb’s formula to solve for \(r\). This includes calculating the square root when \(r^2\) is determined.
  • In the context of the exercise, knowing how to compute \(r\) helps find the point where forces meet given conditions, combining practical math skills with physics principles.
Understanding this concept is critical for solving problems related to forces involving charged particles. It also helps visualize how changes in spacing between objects affect the interactions of these fundamental forces.

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Most popular questions from this chapter

The cell membrane in a nerve cell has a thickness of \(0.12 \mu \mathrm{m}\). (a) Approximating the cell membrane as a parallel-plate capacitor with a surface charge density of \(5.9 \times 10^{-6} \mathrm{C} / \mathrm{m}^{2}\), find the electric field within the membrane. (b) If the thickness of the membrane were doubled, would your answer to part (a) increase, decrease, or stay the same? Explain.

In the operating room, technicians and doctors must take care not to create an electric spark, since the presence of the oxygen gas used during an operation increases the risk of a a deadly fire. Should the operating-room personnel wear shoes that are conducting or nonconducting? Explain.

When a nerve impulse propagates along a nerve cell, the electric field within the cell membrane changes from \(7.0 \times 10^{5} \mathrm{N} / \mathrm{C}\) in one direction to \(3.0 \times 10^{5} \mathrm{N} / \mathrm{C}\) in the other direction. Approximating the cell membrane as a parallel-plate capacitor, find the magnitude of the change in charge density on the walls of the cell membrane.

The electric field at the point \(x=5.00 \mathrm{cm}\) and \(y=0\) points in the positive \(x\) direction with a magnitude of \(10.0 \mathrm{N} / \mathrm{C} .\) At the point \(x=10.0 \mathrm{cm}\) and \(y=0\) the electric field points in the positive \(x\) direction with a magnitude of \(15.0 \mathrm{N} / \mathrm{C}\). Assuming this electric field is produced by a single point charge, find (a) its location and (b) the sign and magnitude of its charge.

A point charge \(q=-0.35 \mathrm{nC}\) is fixed at the origin. Where must a proton be placed in order for the electric force acting on it to be exactly opposite to its weight? (Let the \(y\) axis be vertical and the \(x\) axis be horizontal.)

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