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Chapter 18: Problem 28

(a) You find that if you place charges of \(\pm 1.25 \mu \mathrm{C}\) on two separated metal objects, the potential difference between them is \(11.3 \mathrm{~V}\). What is their capacitance? (b) A capacitor has a capacitance of \(7.28 \mu \mathrm{F}\). What amount of excess charge must be placed on each of its plates to make the potential difference between the plates equal to \(25.0 \mathrm{~V} ?\)

Short Answer

Expert verified
Capacitance for part (a) is approximately 111 nF. The excess charge for part (b) is 182 碌C.

Step by step solution

01

Understanding Capacitance Formula

The capacitance \( C \) of a capacitor is defined by the formula \( C = \frac{Q}{V} \), where \( Q \) is the charge in coulombs and \( V \) is the potential difference in volts between the two conductors.
02

Calculating Capacitance for Part (a)

In part (a), the charge \( Q = 1.25 \times 10^{-6} \mathrm{~C} \) (since 1 microcoulomb = \( 10^{-6} \mathrm{~C} \)) and the potential difference \( V = 11.3 \mathrm{~V} \). Plug these values into the formula: \( C = \frac{1.25 \times 10^{-6}}{11.3} \) to find the capacitance \( C \).
03

Solving the Capacitance Equation for Part (a)

Perform the calculation \( C = \frac{1.25 \times 10^{-6}}{11.3} \) to determine \( C \). The resulting capacitance is approximately \( 1.11 \times 10^{-7} \mathrm{~F} \) or \( 111 \mathrm{~nF} \) (since 1 nanofarad = \( 10^{-9} \mathrm{~F} \)).
04

Understanding Required Charge Calculation for Part (b)

For part (b), rearrange the capacitance formula to find the charge: \( Q = C \times V \). Here, the capacitance \( C = 7.28 \times 10^{-6} \mathrm{~F} \) and potential difference \( V = 25.0 \mathrm{~V} \).
05

Calculating Excess Charge for Part (b)

Substitute the given values into the equation: \( Q = 7.28 \times 10^{-6} \times 25.0 \) to find the excess charge \( Q \).
06

Solving the Charge Equation for Part (b)

Perform the calculation \( Q = 7.28 \times 10^{-6} \times 25.0 \). The excess charge \( Q \) is approximately \( 1.82 \times 10^{-4} \mathrm{~C} \) or \( 182 \mathrm{~ ext{碌C}} \).

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Key Concepts

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

Potential Difference
Potential difference, also known as voltage, is the measure of the electrical potential energy difference between two points in an electric field. It is denoted by the symbol \( V \). Potential difference is the driving force that causes electric charges to flow in a circuit, similar to how a difference in height causes water to flow down a slope. In the context of capacitors:
  • The potential difference is the voltage difference between the two plates of a capacitor.
  • This difference is created by separating equal and opposite charges on the plates, leading to an electric field.
  • A higher potential difference means a stronger electric field and a larger separation of charge.
Measuring potential difference is crucial when dealing with circuits and electronic components, as it dictates how much energy per charge is available. For instance, in the exercise, the potential difference was used to calculate capacitance using the given charge.
Electric Charge
Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electromagnetic field. Charge is quantized, meaning it comes in discrete amounts, and the unit of charge is the coulomb (\( C \)). Key points about electric charge:
  • Charges can be positive or negative, often carried by protons and electrons respectively.
  • Like charges repel each other, while opposite charges attract.
  • Charging an object involves moving charges to or from the object, creating an imbalance.
In capacitors, electric charge is stored on the plates, creating an electric field between them. The stored charge directly impacts the potential difference across the capacitor. In question (a), knowing the charge on the objects helps in calculating the capacitance by using the relationship between charge and voltage.
Capacitors
Capacitors are passive electrical components that store energy in the form of an electric field, created between a pair of conductors on which equal but opposite electric charges have been placed. Here鈥檚 how capacitors function:
  • When connected to a power source, capacitors accumulate charge on their plates.
  • This storage of charge directly creates a potential difference across the plates.
  • Once fully charged, a capacitor can then release the stored energy into the circuit when needed.
Capacitors are used in various applications such as filtering electronic signals, tuning radios, and in energy storage devices like camera flashes. In part (b) of the exercise, the given capacitance allowed calculation of the extra charge required to reach a specified potential difference.
Capacitance Formula
The capacitance formula is vital for understanding how capacitors behave under different conditions. Mathematically, capacitance \( C \) is expressed as \( C = \frac{Q}{V} \), where \( Q \) is the charge stored in the capacitor, and \( V \) is the potential difference across the plates. Important aspects include:
  • Capacitance is measured in farads (\( F \)), which indicates the capacity to hold charge per unit voltage.
  • It is determined by the physical properties of the capacitor such as the area of the plates and the distance between them.
  • Higher capacitance means more charge stored for the same potential difference.
In the exercises given, this formula is pivotal for computing the capacitance and understanding the amount of charge needed for a particular voltage. The correct application of this formula aids in solving practical problems related to electric circuits.

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Most popular questions from this chapter

A uniform electric field has magnitude \(E\) and is directed in the negative \(x\) direction. The potential difference between point \(a\) (at \(x=0.60 \mathrm{~m}\) ) and point \(b\) (at \(x=0.90 \mathrm{~m}\) ) is \(240 \mathrm{~V}\). (a) Which point, \(a\) or \(b\), is at the higher potential? (b) Calculate the value of \(E\). (c) A negative point charge \(q=-0.200 \mu \mathrm{C}\) is moved from \(b\) to \(a\). Calculate the work done on the point charge by the electric field.

DATA A positive point charge \(Q\) is placed at a position \(x_{0}\) on the \(x\) axis. The potential \(V\) is then measured at various points along the \(x\) axis. The resulting data are given in the table: $$ \begin{array}{ll} \hline x \text { position (m) } & \text { Potential }(V) \\ \hline 0 & 90 \\ 0.5 & 45 \\ 1.5 & 23 \\ 2.0 & 18 \\ 2.5 & 15 \\ \hline \end{array} $$ Make a plot of \(1 / V\) as a function of the position. From this plot, determine the position and magnitude of the charge.

A parallel-plate capacitor \(C\) is charged up to a potential \(V_{0}\) with a charge of magnitude \(Q_{0}\) on each plate. It is then disconnected from the battery, and the plates are pulled apart to twice their original separation. (a) What is the new capacitance in terms of \(C ?\) (b) How much charge is now on the plates in terms of \(Q_{0} ?\) (c) What is the potential difference across the plates in terms of \(V_{0} ?\)

The paper dielectric in a paper-and-foil capacitor is \(0.0800 \mathrm{~mm}\) thick. Its dielectric constant is \(2.50,\) and its dielectric strength is \(50.0 \mathrm{MV} / \mathrm{m} .\) Assume that the geometry is that of a parallel- plate capacitor, with the metal foil serving as the plates. (a) What area of each plate is required for a \(0.200 \mu \mathrm{F}\) capacitor? (b) If the electric field in the paper is not to exceed one-half the dielectric strength, what is the maximum potential difference that can be applied across the capacitor?

Electrical sensitivity of sharks. Certain sharks can detect an electric field as weak as \(1.0 \mu \mathrm{V} / \mathrm{m}\). To grasp how weak this field is, if you wanted to produce it between two parallel metal plates by connecting an ordinary \(1.5 \mathrm{~V}\) AA battery across these plates, how far apart would the plates have to be?
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