Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 81

Capacitors are used in many applications where you need to supply a short burst of energy. A \(100.0-\mu \mathrm{F}\) capacitor in an electronic flash lamp supplies an average power of \(10.0 \mathrm{kW}\) to the lamp for $2.0 \mathrm{ms}$. (a) To what potential difference must the capacitor initially be charged? (b) What is its initial charge?

Short Answer

Expert verified
Answer: The initial potential difference is 20 V, and the initial charge is 0.002 C or 2.0 mC.

Step by step solution

01

Determine the Initial Potential Difference

First, let's find the initial potential difference (V) using the average power equation. Average power (P_avg) = 0.5 * C * V^2 / time We are given: C = 100.0 μF, P_avg = 10.0 kW, and time = 2.0 ms. To solve for V, we rearrange the formula: V^2 = (2 * P_avg * time) / C Now, plug in the given values and convert units: V^2 = (2 * (10.0 * 10^3 W) * (2.0 * 10^-3 s)) / (100.0 * 10^-6 F) V^2 = (20,000 * 0.002) / 0.0001 V^2 = 400 Take the square root of both sides: V = 20 V So, the initial potential difference is 20 V.
02

Determine the Initial Charge

Next, let's find the initial charge (Q) using the formula: Charge (Q) = Capacitance (C) * Potential difference (V) We are given: C = 100.0 μF and V = 20 V. Plug in the given values and convert units: Q = (100.0 * 10^-6 F) * (20 V) Q = 0.002 C So, the initial charge is 0.002 C or 2.0 mC.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 alpha particle (charge \(+2 e\) ) moves through a potential difference \(\Delta V=-0.50 \mathrm{kV} .\) Its initial kinetic energy is $1.20 \times 10^{-16} \mathrm{J} .$ What is its final kinetic energy?
Charges of \(-12.0 \mathrm{nC}\) and \(-22.0 \mathrm{nC}\) are separated by $0.700 \mathrm{m} .$ What is the potential midway between the two charges?
An electron (charge \(-e\) ) is projected horizontally into the space between two oppositely charged parallel plates. The electric field between the plates is \(500.0 \mathrm{N} / \mathrm{C}\) upward. If the vertical deflection of the electron as it leaves the plates has magnitude \(3.0 \mathrm{mm},\) how much has its kinetic energy increased due to the electric field? [Hint: First find the potential difference through which the electron moves.]
It is believed that a large electric fish known as Torpedo occidentalis uses electricity to shock its victims. A typical fish can deliver a potential difference of \(0.20 \mathrm{kV}\) for a duration of \(1.5 \mathrm{ms}\). This pulse delivers charge at a rate of \(18 \mathrm{C} / \mathrm{s} .\) (a) What is the rate at which work is done by the electric organs during a pulse? (b) What is the total amount of work done during one pulse?
In a region where there is an electric field, the electric forces do $+8.0 \times 10^{-19} \mathrm{J}\( of work on an electron as it moves from point \)X$ to point \(Y\). (a) Which point, \(X\) or \(Y\), is at a higher potential? (b) What is the potential difference, \(V_{Y}-V_{X},\) between point \(Y\) and point \(X ?\)
See all solutions

Recommended explanations on Physics Textbooks

Thermodynamics

Read Explanation

Radiation

Read Explanation

Fundamentals of Physics

Read Explanation

Oscillations

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Torque and Rotational Motion

Read Explanation
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.

Sign-up for free