/*! 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 61 The work done by an electric for... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The work done by an electric force in moving a charge from point \(A\) to point \(B\) is \(2.70 \times 10^{-3}\) J. The electric potential difference between the two points is \(V_{A}-V_{B}=50.0 \mathrm{V} .\) What is the charge?

Short Answer

Expert verified
The charge is \(5.40 \times 10^{-5}\) C.

Step by step solution

01

Defining the Given Information

We are given: 1) The work done by the electric force, which is \(W = 2.70 \times 10^{-3}\) J and 2) The electric potential difference, which is \(V_A - V_B = 50.0\) V. We need to find the charge \(q\).
02

Understanding the Relationship

The work done by the electric force is related to both the electric potential difference and the charge by the formula \(W = q \times (V_A - V_B)\).
03

Rearranging the Formula

To find the charge \(q\), we rearrange the formula to solve for \(q\): \(q = \frac{W}{V_A - V_B}\).
04

Substituting the Values

Plug the given values into the rearranged formula: \(q = \frac{2.70 \times 10^{-3}}{50.0}\) to find the magnitude of the charge.
05

Calculating the Charge

By performing the division, we calculate \(q = \frac{2.70 \times 10^{-3}}{50.0} = 5.40 \times 10^{-5}\) C.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Electric Force
Electric force plays a crucial role in moving charged particles. It arises from electric fields which are created by electrical charges. When a charged object is placed within an electric field, it experiences a force known as the electric force.
Electric force can be thought of as the "push" or "pull" on a charge due to another charge. Specifically, it is proportional to the product of the magnitudes of the charges involved, and inversely proportional to the square of the distance between them. This relationship is described by Coulomb's Law formula:
  • \( F = k \frac{|q_1 \, q_2|}{r^2} \)
where \( F \) is the magnitude of the force, \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between the centers of the two charges.
The direction of this force is along the line joining the two charges. It is attractive if the charges are opposite and repulsive if they are alike.
Work Done
Work done in the context of electric fields refers to the effort required to move a charge against the electric force. This is integral to understanding energy changes in electrical systems.
The mathematical definition of work done by an electric force when moving a charge \( q \) through an electric potential difference \( \Delta V \) is:
  • \( W = q \times (V_A - V_B) \)
It is important to note that the work done depends on three main factors:
  • The electric potential difference \( V_A - V_B \)
  • The amount of charge \( q \)
  • The path taken between the two points
The work done can be positive or negative depending on the direction of the force relative to the movement of the charge.
Charge Calculation
Charge calculation is essential for numerous applications in physics and engineering, especially when analyzing electric circuits and fields.
In this task, the charge is found using the relationship between work done by the electric force and the electric potential difference. By rearranging the formula \( W = q \times (V_A - V_B) \), we solve for the charge:
  • \( q = \frac{W}{V_A - V_B} \)
Substituting the given values in the exercise yields:
  • \( q = \frac{2.70 \times 10^{-3}}{50.0} = 5.40 \times 10^{-5} \, C \)
This calculation demonstrates how to determine the amount of charge, emphasizing the importance of understanding these relationships.
Electric Charge
Electric charge is a fundamental property of matter, representing how much an object will repel or attract other charged objects. It is quantified in coulombs (C).
There are two types of electric charges: positive and negative. These charges are responsible for generating and experiencing electric forces.
Key properties of electric charge include:
  • Conservation: The total charge within an isolated system remains constant.
  • Quantization: Charge exists in discrete amounts, often in multiples of the elementary charge \( e \), where \( e = 1.6 \times 10^{-19} \, C \).
  • Interaction: Like charges repel, while opposite charges attract each other.
Understanding the concept of electric charge is essential in many fields, including electronics, electromagnetism, and chemistry.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An empty parallel plate capacitor is connected between the terminals of a \(9.0-\mathrm{V}\) battery and charged up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor?

The same voltage is applied between the plates of two different capacitors. When used with capacitor A, this voltage causes the capacitor to store \(11 \mu \mathrm{C}\) of charge and \(5.0 \times 10^{-5} \mathrm{J}\) of energy. When used with capacitor \(\mathrm{B}\) which has a capacitance of \(6.7 \mu \mathrm{F}\), this voltage causes the capacitor to store a charge that has a magnitude of \(q_{\mathrm{B}} .\) Determine \(q_{\mathrm{B}}\).

Suppose that the electric potential outside a living cell is higher than that inside the cell by 0.070 V. How much work is done by the electric force when a sodium ion (charge \(=+e\) ) moves from the outside to the inside?

A charge of \(-3.00 \mu \mathrm{C}\) is fixed in place. From a horizontal distance of \(0.0450 \mathrm{m},\) a particle of \(\operatorname{mass} 7.20 \times 10^{-3} \mathrm{kg}\) and charge \(-8.00 \mu \mathrm{C}\) is fired with an initial speed of \(65.0 \mathrm{m} / \mathrm{s}\) directly toward the fixed charge. How far does the particle travel before its speed is zero?

The inner and outer surfaces of a cell membrane carry a negative and a positive charge, respectively. Because of these charges, a potential difference of about \(0.070 \mathrm{V}\) exists across the membrane. The thickness of the cell membrane is \(8.0 \times 10^{-9} \mathrm{m} .\) What is the magnitude of the electric field in the membrane?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.