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Chapter 18: Problem 67

A parallel-plate air capacitor has a capacitance of 920 pF. The charge on each plate is 2.55\(\mu \mathrm{C}\) . (a) What is the potential difference between the plates? (b) If the charge is kept constant, what will be the potential difference between the plates if the separation is doubled? (c) How much work is required to double the separation?

Short Answer

Expert verified
a) 2770 V, b) 5540 V, c) 3.23 mJ of work is needed.

Step by step solution

01

Identify the Formula for Potential Difference

To find the potential difference (V) between the plates of a capacitor, use the formula: \( V = \frac{Q}{C} \), where \( Q \) is the charge on the plates and \( C \) is the capacitance.
02

Calculate Potential Difference (Part a)

Given \( Q = 2.55 \, \mu\text{C} = 2.55 \times 10^{-6} \text{C} \) and \( C = 920 \, \text{pF} = 920 \times 10^{-12} \text{F} \), use the formula: \( V = \frac{2.55 \times 10^{-6}}{920 \times 10^{-12}} = 2.77 \times 10^3 \, \text{V} \).
03

Understand the Effect of Separation Change

If the separation is doubled while keeping the charge constant, the capacitance will become half, because capacitance is inversely proportional to separation \( (C \propto \frac{1}{d}) \).
04

Calculate New Capacitance (Part b)

The new capacitance \( C' \) is \( \frac{C}{2} = \frac{920 \times 10^{-12}}{2} = 460 \times 10^{-12} \text{F} \).
05

Find New Potential Difference (Part b)

With the same charge \( Q = 2.55 \times 10^{-6} \text{C} \) but the new capacitance \( 460 \times 10^{-12} \text{F} \), the new potential difference \( V' = \frac{2.55 \times 10^{-6}}{460 \times 10^{-12}} = 5.54 \times 10^3 \, \text{V} \).
06

Calculate Work Done to Change Separation (Part c)

The work done to change the separation is equal to the change in the energy stored: \( W = \Delta U = \frac{1}{2}Q^2\left(\frac{1}{C'} - \frac{1}{C}\right) \).
07

Compute Work Done (Part c)

Using \( Q = 2.55 \times 10^{-6} \text{C} \), \( C = 920 \times 10^{-12} \text{F} \), and \( C' = 460 \times 10^{-12} \text{F} \), calculate \( W = \frac{1}{2} \times \frac{(2.55 \times 10^{-6})^2}{460 \times 10^{-12}} - \frac{1}{2} \times \frac{(2.55 \times 10^{-6})^2}{920 \times 10^{-12}} = 3.23 \times 10^{-3} \, \text{J} \).

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Key Concepts

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

Potential Difference
The potential difference between the plates of a capacitor, often denoted as \( V \), is an essential concept in understanding how capacitors function. This potential difference is the amount of electric potential energy per unit charge that is needed to move a charge between the plates of a capacitor. It is calculated using the formula \( V = \frac{Q}{C} \) where:
  • \( Q \) is the charge stored on one plate of the capacitor in Coulombs.
  • \( C \) is the capacitance in Farads, describing the capacitor's ability to store charge.
If a given capacitor holds a charge of 2.55 microcoulombs and has a capacitance of 920 picofarads, we can compute the potential difference between the plates by substituting these values into the formula, giving us \( 2.77 \times 10^3 \) volts.
By changing the parameters, we see how the potential difference is affected, bringing deeper insights into capacitor behavior.
Capacitor Charge
Charge in a capacitor is denoted by \( Q \) and is one of the most fundamental attributes of a capacitor. It is the measure of electric charge stored on the plates. In this problem, the charge is given as 2.55 microcoulombs, which equals \( 2.55 \times 10^{-6} \) Coulombs.
When the separation distance between the plates of the capacitor is altered while maintaining the charge, the potential difference changes. This modification does not change the stored charge itself but impacts how that charge interacts with the changing electric field.
Thus, charge stability is important in examining how different physical changes to the capacitor influence other properties like capacitance and potential difference.
Electric Field
The electric field inside a capacitor is a constant field that develops between the two charged plates. It is more intense when the charges are closer together and less intense as they are further separated. The electric field in a parallel-plate capacitor can be considered uniform, pointing from the positive plate to the negative one.
  • It is related to potential difference and separation by \( E = \frac{V}{d} \), where \( d \) is the separation distance between plates.
  • The stronger the electric field, the more potential energy per charge available in the system.
When you change the distance between the plates, things like doubling the distance lead to a decrease in the electric field strength, affecting other parameters like capacitance, serving in demonstrating the interdependence of these core concepts.
Energy Stored in Capacitor
A capacitor not only stores charge but also energy, commonly known as the electric potential energy. This energy is stored in the electric field between the plates. The energy stored \( U \) in a capacitor can be expressed in multiple formulas, with one being \( U = \frac{1}{2} C V^2 \).
In this provided problem, another formula is applied during the calculation of work done to change the separation, \( W = \Delta U = \frac{1}{2}Q^2\left(\frac{1}{C'} - \frac{1}{C}\right) \). This equation highlights how altering plate separation impacts stored energy and requires work due to changes in capacitance and potential difference.
  • Stored energy provides insights into the potential capacity of the circuit to perform work.
  • Understanding energy transformations in capacitors underlines their significance in circuits, such as energy management, regulation, and storage.

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Most popular questions from this chapter

In the text, it was shown that the energy stored in a capacitor \(C\) charged to a potential \(V\) is \(U=\frac{1}{2} Q V\) . Show that this energy can also be expressed as (a) \(U=Q^{2} / 2 C\) and (b) \(U=\frac{1}{2} C V^{2}\)

Capacitance of an oscilloscope. Oscilloscopes have parallel metal plates inside them to deflect the electron beam. These plates are called the deflecting plates. Typically, they are squares 3.0 \(\mathrm{cm}\) on a side and separated by \(5.0 \mathrm{mm},\) with vacuum in between. What is the capacitance of these deflecting plates and hence of the oscilloscope? (This capacitance can sometimes have an effect on the circuit you are trying to study and must be taken into consideration in your calculations.)

(a) An electron is to be accelerated from \(3.00 \times 10^{6} \mathrm{m} / \mathrm{s}\right.\) to \(8.00 \times 10^{6} \mathrm{m} / \mathrm{s}\) . Through what potential difference must the electron pass to accomplish this? (b) Through what potential difference must the electron pass if it is to be slowed from \(8.00 \times 10^{6} \mathrm{m} / \mathrm{s}\) to a halt?

Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit, the flash lasts for \(\frac{1}{675}\) s with an average light power output of \(2.70 \times 10^{5} \mathrm{W}\) (a) If the conversion of electrical energy to light is 95\(\%\) efficient (the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash? (b) The capacitor has a potential difference between its plates of 125 \(\mathrm{V}\) when the stored energy equals the value calculated in part (a). What is the capacitance?

A 10.0\(\mu \mathrm{F}\) parallel-plate capacitor with circular plates is connected to a 12.0 \(\mathrm{V}\) battery. (a) What is the charge on each plate? (b) How much charge would be on the plates if their separation were doubled while the capacitor remained connected to the battery? (c) How much charge would be on the plates if the capacitor were connected to the 12.0 \(\mathrm{V}\) battery after the radius of each plate was doubled without changing their separation?
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