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Chapter 18: Problem 51

(a) How much charge does a battery have to supply to a \(5.0 \mu \mathrm{F}\) capacitor to create a potential difference of \(1.5 \mathrm{~V}\) across its plates? How much energy is stored in the capacitor in this case? (b) How much charge would the battery have to supply to store \(1.0 \mathrm{~J}\) of energy in the capacitor? What would be the potential across the capacitor in that case?

Short Answer

Expert verified
(a) Charge: 7.5 碌C; Energy: 5.625 碌J. (b) Charge: 3.1625 mC; Potential: 632.5 V.

Step by step solution

01

Understanding the problem

We are given a capacitor with a capacitance of \(5.0 \mu\text{F}\) and need to find two things for part (a): the charge supplied by the battery to create a potential difference of \(1.5 \,\text{V}\) across it, and the energy stored in the capacitor. For part (b), we need to determine the charge required to store \(1.0 \,\text{J}\) of energy and the resulting potential difference.
02

Formula for charge on a capacitor

The formula to find the charge \( Q \) stored on a capacitor is given by: \[ Q = C \times V \] where \( C \) is the capacitance and \( V \) is the potential difference.
03

Calculating charge for part (a)

Using the formula from Step 2, substituting \( C = 5.0 \times 10^{-6} \,\text{F} \) and \( V = 1.5 \,\text{V} \): \[ Q = 5.0 \times 10^{-6} \,\text{F} \times 1.5 \,\text{V} = 7.5 \times 10^{-6} \text{C} \] The charge supplied by the battery is \(7.5 \,\mu\text{C}\).
04

Formula for energy stored in a capacitor

The energy \( E \) stored in a capacitor is given by the formula: \[ E = \frac{1}{2} C V^2 \] where \( C \) is the capacitance and \( V \) is the potential difference.
05

Calculating energy for part (a)

Substituting the values into the energy formula: \[ E = \frac{1}{2} \times 5.0 \times 10^{-6} \,\text{F} \times (1.5 \,\text{V})^2 = 5.625 \times 10^{-6} \text{J} \] The energy stored in the capacitor is approximately \(5.625 \,\mu\text{J}\).
06

Formula for energy to find charge for part (b)

We use the energy formula again: \[ E = \frac{1}{2} C V^2 \] We need \( E = 1.0 \,\text{J} \). We will find \( Q \) using \( Q = C \times V \). Rearrange the formula to express \( V \) in terms of \( Q \) and \( C \).
07

Solving for potential \( V \) using energy for part (b)

Rearrange the formula to solve for \( V^2 \): \[ V = \sqrt{\frac{2E}{C}} \] Substituting \( E = 1.0 \,\text{J} \) and \( C = 5.0 \times 10^{-6} \,\text{F} \): \[ V = \sqrt{\frac{2 \times 1.0 \,\text{J}}{5.0 \times 10^{-6} \,\text{F}}} = \sqrt{400000} = 632.5 \,\text{V} \]
08

Calculating charge for part (b)

Finally, compute the charge using \( C \times V \): \[ Q = 5.0 \times 10^{-6} \,\text{F} \times 632.5 \,\text{V} = 3.1625 \times 10^{-3} \text{C} \] The charge needed to store \(1.0 \,\text{J}\) is approximately \(3.1625 \,\text{mC}\).

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Key Concepts

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

Charge Calculation
A capacitor stores electrical energy, and this energy is related to the charge stored. If you have a capacitor with a given capacitance and you know the potential difference you want to establish across it, you can calculate the amount of charge the battery must supply. The formula used for this calculation is:
  • \( Q = C \times V \)
where \( Q \) is the charge, \( C \) is the capacitance in farads, and \( V \) is the potential difference in volts. By multiplying the capacitance and the potential difference, we can determine the charge. This formula tells us that the stored charge is directly proportional to both the capacitance and the voltage.
Potential Difference
The potential difference, or voltage, across a capacitor is critical for determining how much energy it can store. When you need to figure out the voltage across a capacitor, especially when a specific amount of energy is stored, another formula ties the energy, capacitance, and voltage together.
  • \( E = \frac{1}{2} C V^2 \)
  • \( V = \sqrt{\frac{2E}{C}} \)
This means that for a given energy, the voltage will be higher if the capacitance is smaller. Conversely, if capacitance is larger, a lower voltage would yield the same energy storage. Understanding this relationship is crucial when designing circuits that require precise voltage settings.
Energy Stored in a Capacitor
Capacitors store energy in the form of an electric field. The energy stored (\( E \)) can be calculated through the formula:
  • \( E = \frac{1}{2} C V^2 \)
where \( C \) is the capacitance and \( V \) is the voltage across the capacitor. This equation is derived from how capacitors accumulate charge, and it shows that the energy stored is proportional to the square of the voltage and directly proportional to the capacitance. This quadratic relationship means even small increases in voltage lead to significant increases in stored energy.
Capacitance
Capacitance is a measure of a capacitor's ability to store charge per unit voltage. It is inherently dependent on the physical attributes of the capacitor, such as the size of its plates and the separation between them, as well as the material between the plates (the dielectric). Mathematically, it is expressed as:
  • \( C = \frac{Q}{V} \)
where \( Q \) is the charge stored and \( V \) is the voltage across the capacitor. High capacitance means more charge storage for the same voltage, important in many electronic applications. Capacitors with larger capacitance are often used to smooth out fluctuations in power supply lines and in power conditioning.

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Most popular questions from this chapter

How much charge does a \(12 \mathrm{~V}\) battery have to supply to fully charge a \(2.5 \mu \mathrm{F}\) capacitor and a \(5.0 \mu \mathrm{F}\) capacitor when they're (a) in parallel, (b) in series? (c) How much energy does the battery have to supply in each case?

Cell membranes. Cell membranes (the walled enclosure around a cell) are typically about \(7.5 \mathrm{nm}\) thick. They are partially permeable to allow charged material to pass in and out, as needed. Equal but opposite charge densities build up on the inside and outside faces of such a membrane, and these charges prevent additional charges from passing through the cell wall. We can model a cell membrane as a parallel-plate capacitor, with the membrane itself containing proteins embedded in an organic material to give the membrane a dielectric constant of about 10\. (See Figure \(18.47 .)\) (a) What is the capacitance per square centimeter of such a cell wall? (b) In its normal resting state, a cell has a potential difference of \(85 \mathrm{mV}\) across its membrane. What is the electric field inside this membrane?

Two protons are released from rest when they are \(0.750 \mathrm{nm}\) apart. (a) What is the maximum speed they will reach? When does this speed occur? (b) What is the maximum acceleration they will achieve? When does this acceleration occur?

Electric eels and electric fish generate large potential differences that are used to stun enemies and prey. These potentials are produced by cells that each can generate \(0.10 \mathrm{~V}\). We can plausibly model such cells as charged capacitors. (a) How should these cells be connected (in series or in parallel) to produce a total potential of more than \(0.10 \mathrm{~V} ?\) (b) Using the connection in part (a), how many cells must be connected together to produce the \(500 \mathrm{~V}\) surge of the electric eel?

At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are \(4.98 \mathrm{~V}\) and \(12.0 \mathrm{~V} / \mathrm{m},\) respectively. (Take the potential to be zero at infinity.) (a) What is the distance to the point charge? (b) What is the magnitude of the charge? (c) Is the electric field directed toward or away from the point charge?
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