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Interactive LearningWare 19.2 at reviews the concepts pertinent to this problem. What is the potential difference between the plates of a 3.3-F capacitor that stores sufficient energy to operate a 75 -W light bulb for one minute?

Short Answer

Expert verified
The potential difference is approximately 52.22 V.

Step by step solution

01

Understand the Problem

We need to find the potential difference across the plates of a capacitor, given the capacitance and the energy needed to operate a light bulb for a specific time.
02

Gather the Given Information

We are given a capacitance of the capacitor, \( C = 3.3 \, \text{F} \), and we need the capacitor to store energy for a \( 75 \, \text{W} \) light bulb for one minute. Power \( P = 75 \, \text{W} \), and time \( t = 60 \, \text{s} \).
03

Calculate the Energy Required

Energy can be calculated using the power formula: \( E = P \times t \). Substitute the given values: \( E = 75 \, \text{W} \times 60 \, \text{s} = 4500 \, \text{J} \).
04

Use the Energy Stored in Capacitor Formula

The energy stored in a capacitor is given by \( E = \frac{1}{2} C V^2 \) where \( V \) is the potential difference. Rearrange to find \( V \): \( V = \sqrt{\frac{2E}{C}} \).
05

Substitute Values to Find Voltage

Substitute \( E = 4500 \, \text{J} \) and \( C = 3.3 \, \text{F} \) into the formula: \( V = \sqrt{\frac{2 \times 4500}{3.3}} \approx \sqrt{2727.27} \approx 52.22 \, \text{V} \).

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Key Concepts

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

Capacitor Energy Storage
A capacitor is a device that can store electrical energy within its electric field. This stored energy is what allows capacitors to play a critical role in electronic circuits and systems. When a capacitor accumulates charge on its plates, it stores this energy in the form of an electric field between the plates. The ability to store energy makes capacitors useful for applications like smoothing the supply of power in electronic devices and delivering rapid bursts of energy.

The energy stored in a capacitor can be calculated using the formula:
  • \(E = \frac{1}{2} C V^2\)
Here, \(E\) stands for the energy in joules (J), \(C\) is the capacitance in farads (F), and \(V\) is the potential difference in volts (V). This formula shows that the energy stored in a capacitor is proportional to the square of the voltage and directly proportional to the capacitance.

For instance, in the context of powering a light bulb, the energy stored within the capacitor is critical as it determines how long the device can operate. Understanding this concept is fundamental when designing circuits that are reliant on capacitors for their power requirements.
Power and Energy Calculations
Calculating power and energy is central to understanding how much work can be done by a system. Power, which is measured in watts (W), quantifies the rate at which energy is consumed or produced within a system. When faced with practical scenarios, like lighting a bulb, understanding power and energy calculations helps us determine how long a certain power level can be sustained.

The formula for calculating energy given power and time is:
  • \(E = P \times t\)
Here, \(E\) represents energy in joules (J), \(P\) is power in watts (W), and \(t\) is time in seconds (s). In our exercise, the given problem involves supplying a \(75\,\mathrm{W}\) light bulb with energy over a span of one minute. Using the formula, we calculate the energy required as \(4500\,\mathrm{J}\).

This fundamental concept of power and energy helps in evaluating the efficiency and feasibility of energy systems and their components. By understanding this, you can predict how long devices can operate given a specific power source, maximizing efficiency in various applications.
Capacitor Formulas
Various capacitor formulas are crucial for solving problems related to energy storage and potential difference across capacitors. These formulas allow us to predict and calculate different parameters necessary for the design and analysis of circuits.

One of the essential formulas is the energy stored in a capacitor, which is given by:
  • \(E = \frac{1}{2} C V^2\)
This directly relates stored energy with capacitance and voltage. Another key formula relates to the potential difference across the plates of a capacitor, which can be derived as:
  • \(V = \sqrt{\frac{2E}{C}}\)
In this context, we calculate the voltage needed to store a certain amount of energy, like the 4500 J required to power a light bulb for a minute. With a capacitance of \(3.3\,\mathrm{F}\), the potential difference comes out to approximately \(52.22\,\mathrm{V}\).

Understanding these formulas allows engineers and students to design and analyze circuits more effectively, ensuring that components operate within their intended parameters. They form the foundation for many more complex electric phenomena observed in engineering and physics.

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Most popular questions from this chapter

Two identical point charges are fixed to diagonally opposite corners of a square that is \(0.500 \mathrm{~m}\) on a side. Each charge is \(+3.0 \times 10^{-6} \mathrm{C} .\) How much work is done by the electric force as one of the charges moves to an empty corner?

Multiple-Concept Example 4 provides useful background for this problem. Point \(\mathrm{A}\) is at a potential of \(+250 \mathrm{~V}\), and point \(B\) is at a potential of \(-150 \mathrm{~V}\). An \(\alpha\) -particle is a helium nucleus that contains two protons and two neutrons; the neutrons are electrically neutral. An \(\alpha\) -particle starts from rest at \(A\) and accelerates toward \(B\). When the \(\alpha\) -particle arrives at \(B\), what kinetic energy (in electron volts) does it have?

Charges \(q_{1}\) and \(q_{2}\) are fixed in place, \(q_{2}\) being located at a distance \(d\) to the right of \(\mathrm{q}_{1} . \mathrm{A}\) third charge \(q_{3}\) is then fixed to the line joining \(q_{1}\) and \(q_{2}\) at a distance \(d\) to the right of \(q_{2}\). The third charge is chosen so the potential energy of the group is zero; that is, the potential energy has the same value as that of the three charges when they are widely separated. Determine the value for \(q_{3}\), assuming that (a) \(q_{1}=q_{2}=q\) and (b) \(q_{1}=q\) and \(q_{2}=-q .\) Express your answers in terms of \(q\)

When you walk across a rug on a dry day, your body can become electrified, and its electric potential can change. When the potential becomes large enough, a spark of negative charges can jump between your hand and a metal surface. A spark occurs when the electric field strength created by the charges on your body reaches the dielectric strength of the air. The dielectric strength of the air is \(3.0 \times 10^{6} \mathrm{~N} / \mathrm{C}\) and is the electric field strength at which the air suffers electrical breakdown. Suppose a spark \(3.0 \mathrm{~mm}\) long jumps between your hand and a metal doorknob. Assuming that the electric field is uniform, find the potential difference \(\left(V_{\mathrm{knob}}-V_{\text {hand }}\right)\) between your hand and the doorknob.

Location \(A\) is \(3.00 \mathrm{~m}\) to the right of a point charge \(q .\) Location \(B\) lies on the same line and is \(4.00 \mathrm{~m}\) to the right of the charge. The potential difference between the two locations is \(V_{B}-V_{A}=45.0 \mathrm{~V}\). What is the magnitude and sign of the charge?

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