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Chapter 19: Problem 1

During a particular thunderstorm, the electric potential difference between a cloud and the ground is \(V_{\text { clound }}-V_{\text { ground }}=1.3 \times 10^{8} \mathrm{V},\) with the cloud being at the higher potential. What is the change in an electron's elec- tric potential energy when the electron moves from the ground to the cloud?

Short Answer

Expert verified
The change in potential energy is \(-2.08 \times 10^{-11} \text{ J}\).

Step by step solution

01

Understanding the Concept

Electric potential energy changes when a charge moves through a potential difference. The change in electric potential energy (\(\Delta U\)) of a charge (\(q\)) is calculated by multiplying the charge by the potential difference (\(\Delta V\)).
02

Identifying Known Values

We know that the potential difference between the cloud and the ground is \(\Delta V = V_{\text{cloud}} - V_{\text{ground}} = 1.3 \times 10^8 \text{ V}\). The charge of an electron is \(q = -1.6 \times 10^{-19} \text{ C}\).
03

Applying the Formula for Potential Energy Change

The formula for the change in potential energy is \(\Delta U = q \times \Delta V\). We will substitute the values for the electron's charge and the potential difference into this formula.
04

Calculating the Change in Potential Energy

Substitute the known values: \(\Delta U = (-1.6 \times 10^{-19} \text{ C}) \times (1.3 \times 10^8 \text{ V})\).
05

Performing the Calculation

Calculate \(\Delta U = -1.6 \times 1.3 \times 10^{-11} = -2.08 \times 10^{-11} \text{ J}\).
06

Conclusion

The change in the electric potential energy of the electron is \(-2.08 \times 10^{-11} \text{ J}\) as it moves from the ground to the cloud.

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Key Concepts

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

Electric Potential Difference
The electric potential difference, often referred to as voltage, describes the difference in electric potential energy per unit charge between two points. In simple terms, it's like a pressure difference that pushes charges to move from one point to another. If you imagine a hill, the potential difference is similar to the difference in height that causes water to flow downwards. In a thunderstorm situation, the cloud typically has a higher potential than the ground. This potential difference is what energizes electrons, allowing them to move and do work.
  • The potential difference is given in volts (V).
  • A higher potential difference means more energy is available to move a charge.
In this exercise, the potential difference is 1.3 x 10^8 V from the ground to the cloud.
Electron Charge
Electron charge is the electric charge carried by a single electron. It is a fundamental property of electrons and is a basic building block in the study of electricity and magnetism. The charge of an electron is a negative value, denoted by \[q = -1.6 \times 10^{-19} \, \text{Coulombs (C)}.\]This small and negative charge means electrons will move towards a higher potential area within an electric field since opposite charges attract.
  • Electrons are fundamental particles with a consistent charge magnitude.
  • This negative charge makes electrons behave in specific ways when exposed to electric fields and potential differences.
Understanding electron charge is vital in calculating electric potential energy changes, which depend on both the charge value and the potential difference.
Potential Energy Calculation
Potential energy calculation involves understanding how much energy is gained or lost by a charge moving through an electric field. The change in electric potential energy (\(\Delta U\)) when a charge moves through a potential difference (\(\Delta V\)) can be calculated using the formula:\[\Delta U = q \times \Delta V\]where \(q\) is the charge of the particle. This formula tells us how much work needs to be done to move the charge through the potential difference.
  • Substitute the electron's charge and the potential difference into the formula to find the energy change.
  • This calculation helps us understand how electric fields do work on charges, moving them through space.
In the exercise, calculating the energy change involved multiplying the electron charge (\(-1.6 \times 10^{-19}\) C) by the potential difference (\(1.3 \times 10^8\) V), resulting in an energy change of \(-2.08 \times 10^{-11}\) Joules.

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Most popular questions from this chapter

Two identical capacitors store different amounts of energy: capacitor A stores \(3.1 \times 10^{-3} \mathrm{J}\) and capacitor \(\mathrm{B}\) stores \(3.4 \times 10^{-4} \mathrm{J}\) . The voltage across the plates of capacitor \(\mathrm{B}\) is 12 \(\mathrm{V}\) . Find the voltage across the plates of capacitor A.

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 capacitor is constructed of two concentric conducting cylindrical shells. The radius of the inner cylindrical shell is \(2.35 \times 10^{-3} \mathrm{m},\) and that of the outer shell is 2.50 \(\times 10^{-3} \mathrm{m}\) . When the cylinders carry equal and opposite charges of magnitude 1.7 \(\times 10^{-10} \mathrm{C}\) , the electric field between the plates has an average magnitude of \(4.2 \times 10^{4} \mathrm{V} / \mathrm{m}\) and is directed radially outward from the inner shell to the outer shell. Determine (a) the magnitude of the potential difference between the cylindrical shells and \((\text { b) the capacitance of this capacitor. }\)

If the electric field inside a capacitor exceeds the dielectric strength of the dielectric between its plates, the dielectric will break down, discharging and ruining the capacitor. Thus, the dielectric strength is the maximum magnitude that the electric field can have without breakdown occurring. The dielectric strength of air is \(3.0 \times 10^{6} \mathrm{V} / \mathrm{m},\) and that of neoprene rubber is \(1.2 \times 10^{7} \mathrm{V} / \mathrm{m} .\) A certain air-gap, parallel plate capacitor can store no more than 0.075 \(\mathrm{J}\) of electrical energy before breaking down. How much energy can this capacitor store with- out breaking down after the gap between its plates is filled with neoprene rubber?

A particle with a charge of \(-1.5 \mu \mathrm{C}\) and a mass of \(2.5 \times 10^{-6} \mathrm{kg}\) is released from rest at point \(A\) and accelerates toward point \(B\) , arriving there with a speed of 42 \(\mathrm{m} / \mathrm{s}\) . The only force acting on the particle is the electric force. (a) Which point is at the higher potential? Give your reasoning. (b) What is the potential difference \(V_{\mathrm{B}}-V_{\mathrm{A}}\) between \(\mathrm{A}\) and \(\mathrm{B} ?\)
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