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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

Understand the Formula

The work done by an electric force can be calculated using the formula \( W = q \cdot (V_A - V_B) \), where \( W \) is the work done, \( q \) is the charge, and \( V_A - V_B \) is the potential difference between points A and B.
02

Rearrange the Formula

Rearrange the formula to solve for the charge \( q \). This gives us \( q = \frac{W}{V_A - V_B} \).
03

Substitute Known Values

Substitute the given values into the equation: Work \( W = 2.70 \times 10^{-3} \) J and potential difference \( V_A - V_B = 50.0 \) V. So, \( q = \frac{2.70 \times 10^{-3}}{50.0} \).
04

Calculate the Charge

Perform the division to find the charge \( q \). Calculate \( q = \frac{2.70 \times 10^{-3}}{50.0} = 5.40 \times 10^{-5} \) C.

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Key Concepts

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

Work Done by Electric Force
When we talk about work done by an electric force, we are discussing the energy required to move a charge from one point to another in the presence of an electric field. Imagine pushing a shopping cart uphill; the work done is the effort it takes to move it. Similarly, in the world of charges, work is needed to move these tiny particles against the forces within an electric field.

To calculate the work done by an electric force, we use the formula:
  • \( W = q \cdot (V_A - V_B) \)
- Here, \( W \) represents work in joules (J).- \( q \) is the charge, which is what we're solving for.- \( V_A - V_B \) is the electric potential difference between points A and B.

Knowing how much work is done is crucial because it helps us determine the amount of energy used or needed in processes involving charges, such as in circuits or electronic devices.
Electric Potential Difference
The electric potential difference, often dubbed voltage, is a measure of how much potential energy a charge has at one point compared to another. Think of it like the difference in height between two points on a hill; the higher you are, the more potential energy you have.

In terms of electric charges:
  • The potential difference \( V_A - V_B \) tells us how much work is needed to move a unit of charge from point B to point A.
- A higher potential difference means more work is needed to move the charge, similar to climbing a steeper hill.

In the context of the problem, the potential difference is given as 50.0 V. This tells us that a potential energy change of 50.0 joules would occur per coulomb of charge moving between the two points. Understanding potential difference helps us grasp how and why charges move, which is fundamental in technologies like batteries and power supplies.
Electrostatics Formulas
Electrostatics is the study of forces between charges, and it relies on key formulas to describe these interactions. These formulas help us predict how charges behave in an electric field.

Here are some core formulas in electrostatics:
  • **Coulomb's Law:** \( F = k \cdot \frac{|q_1 \cdot q_2|}{r^2} \)
    This formula calculates the force between two charges, where \( F \) is the force, \( k \) is the electrostatic constant, \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between them.
  • **Electric Field:** \( E = \frac{F}{q} = k \cdot \frac{|q|}{r^2} \)
    The electric field \( E \) represents the force per unit charge exerted on a small positive charge in the field.
  • **Work Done by Electric Force:** \( W = q \cdot (V_A - V_B) \)
    This is the formula used in the given problem, linking work, charge, and potential difference.

Understanding these formulas is essential as they form the backbone of electrostatics theory. They provide insights into interactions between charges, how to calculate forces and fields, and lead us to solve practical problems in electric circuits and beyond.

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Most popular questions from this chapter

The electric potential energy stored in the capacitor of a defibrillator is 73 J, and the capacitance is \(120 \mu \mathrm{F}\). What is the potential difference that exists across the capacitor plates?

Equipotential surface \(A\) has a potential of \(5650 \mathrm{V},\) while equipotential surface \(B\) has a potential of 7850 V. A particle has a mass of \(5.00 \times 10^{-2} \mathrm{kg}\) and a charge of \(+4.00 \times 10^{-5} \mathrm{C} .\) The particle has a speed of \(2.00 \mathrm{m} / \mathrm{s}\) on surface \(A .\) A nonconservative outside force is applied to the particle, and it moves to surface \(B\), arriving there with a speed of \(3.00 \mathrm{m} / \mathrm{s}\). How much work is done by the outside force in moving the particle from \(A\) to \(B ?\)

A spark plug in an automobile engine consists of two metal conductors that are separated by a distance of \(0.75 \mathrm{mm} .\) When an electric spark jumps between them, the magnitude of the electric field is \(4.7 \times 10^{7} \mathrm{V} / \mathrm{m} .\) What is the magnitude of the potential difference \(\Delta V\) between the conductors?

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?

A moving particle encounters an external electric field that decreases its kinetic energy from \(9520 \mathrm{eV}\) to \(7060 \mathrm{eV}\) as the particle moves from position \(A\) to position \(B\). The electric potential at \(A\) is \(-55.0 \mathrm{V}\), and the electric potential at \(B\) is \(+27.0 \mathrm{V}\). Determine the charge of the particle. Include the algebraic sign ( \(+\) or \(-\) ) with your answer.

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