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Chapter 24: Problem 3

(I) The potential difference between two short sections of parallel wire in air is 24.0 \(\mathrm{V}\) . They carry equal and opposite charge of magnitude 75 \(\mathrm{pC}\) . What is the capacitance of the two wires?

Short Answer

Expert verified
The capacitance is 3.125 pF.

Step by step solution

01

Understanding the Problem

We are given a potential difference \( V = 24.0 \, \mathrm{V} \) and equal and opposite charges \( Q = 75 \, \mathrm{pC} \) on two parallel wires. We need to find the capacitance \( C \) of this system.
02

Using the Capacitance Formula

The formula for capacitance \( C \) is given by \( C = \frac{Q}{V} \). Here, \( Q = 75 \, \mathrm{pC} = 75 \times 10^{-12} \, \mathrm{C} \) and \( V = 24.0 \, \mathrm{V} \).
03

Substituting the Values

Substitute the given values into the capacitance formula to calculate \( C \):\[ C = \frac{75 \times 10^{-12}}{24.0} \]
04

Calculate the Capacitance

Perform the division to find \( C \):\[ C = \frac{75 \times 10^{-12}}{24.0} = 3.125 \times 10^{-12} \, \mathrm{F} \]
05

Final Result

The capacitance of the two parallel wires is \( 3.125 \, \mathrm{pF} \) (picoFarads).

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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, often called voltage, is a measure of how much energy is available to move charge between two points. Imagine an electric circuit as a water slide: the potential difference is like the height of the slide. The higher the slide, the more energy you gain going down. Similarly, in an electrical system, when you have a higher potential difference, it provides more energy for moving electrons from one place to another.

In our exercise, the potential difference between the two parallel wires is 24 volts. This number indicates the energy per unit charge that drives electrons from one wire to the other or vice versa. It's an essential component in determining how these wires function as a capacitor, which stores energy in the electric field formed between them.
Charge
Charge is a fundamental property of matter that causes it to experience a force when placed in an electromagnetic field. Often measured in coulombs ( C ), it comes in two types: positive and negative. Electrons, for example, carry a negative charge, while protons carry a positive charge.

In the given problem, the parallel wires carry equal and opposite charges of 75 picoCoulombs each. This means each wire holds a tiny amount of electricity, which is crucial for creating an electric field between them. The charge stored in a capacitor directly affects its ability to store energy: the greater the charge for a given potential difference, the greater the capacitance.
Parallel Wires
Parallel wires refer to two conductors with a gap maintaining an electrical potential difference between them. In terms of electrical circuits, this setup is vital for creating capacitors. When two wires are placed parallel to each other, they form a basic form of a capacitor.

The electric field between the wires allows them to store electrical energy. The strength of this setup's capacitance is determined by factors including the distance between the wires, the surface area of the wires facing each other, and the material between them (often air). Challenges in wider applications involve managing these distances accurately and using materials that enhance capacitance without being prohibitive in cost or complexity.

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Most popular questions from this chapter

(II) In a dynamic random access memory (DRAM) computer chip, each memory cell chiefly consists of a capacitor for charge storage. Each of these cells represents a single binary-bit value of 1 when its 35 -fF capacitor \(\left(1 \mathrm{fF}=10^{-15} \mathrm{F}\right)\) is charged at \(1.5 \mathrm{V},\) or 0 when uncharged at 0.V. (a) When it is fully charged, how many excess electrons are on a cell capacitor's negative plate? (b) After charge has been placed on a cell capacitor's plate, it slowly "leaks" off (through a variety of mechanisms) at a constant rate of 0.30 \(\mathrm{fC} / \mathrm{s}\) . How long does it take for the potential difference across this capacitor to decrease by 1.0\(\%\) from its fully charged value? (Because of this leakage effect, the charge on a DRAM capacitor is "refreshed" many times per second.)

A cardiac defibrillator is used to shock a heart that is beating erratically. A capacitor in this device is charged to \(7.5 \mathrm{kV}\) and stores \(1200 \mathrm{~J}\) of energy. What is its capacitance?

The potential difference between two short sections of parallel wire in air is \(24.0 \mathrm{~V}\). They carry equal and opposite charge of magnitude \(75 \mathrm{pC}\). What is the capacitance of the two wires?

(II) \((a)\) Show that each plate of a parallel-plate capacitor exerts a force $$F=\frac{1}{2} \frac{Q^{2}}{\epsilon_{0} A}$$ on the other, by calculating dW/dx where \(d W\) is the work needed to increase the separation by \(d x\) . (b) Why does using \(F=Q E,\) with \(E\) being the electric field between the plates, give the wrong answer?

An electric field of \(4.80 \times 10^{5} \mathrm{~V} / \mathrm{m}\) is desired between two parallel plates, each of area \(21.0 \mathrm{~cm}^{2}\) and separated by \(0.250 \mathrm{~cm}\) of air. What charge must be on each plate?
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