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Chapter 18: Problem 36

(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
(a) 110.6 nF; (b) 182 渭C.

Step by step solution

01

Understanding Capacitance

To find the capacitance, we use the formula \( C = \frac{Q}{V} \), where \( Q \) is the charge and \( V \) is the potential difference. This formula defines capacitance as the charge stored per unit voltage.
02

Calculate Capacitance for Given Charges and Voltage

Given \( Q = 1.25 \mu C = 1.25 \times 10^{-6} C \) and \( V = 11.3 V \), substitute into the formula: \[C = \frac{1.25 \times 10^{-6}}{11.3}\]Calculate to find:\[C \approx 1.106 \times 10^{-7} F = 110.6 nF\]
03

Understanding Charge in Capacitors

For part (b), we need to find the charge \( Q \) needed for a given capacitance and potential difference. Use \( Q = C \times V \).
04

Calculate Charge for Given Capacitance and Voltage

Given \( C = 7.28 \mu F = 7.28 \times 10^{-6} F \) and \( V = 25 V \), substitute into the formula: \[Q = 7.28 \times 10^{-6} \times 25\]Calculate to find:\[Q = 1.82 \times 10^{-4} C = 182 \mu C\]

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Key Concepts

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

Electric Charge
Electric charge is a fundamental property of matter. It can be either positive or negative. Objects can attract or repel each other based on their charges.
When you have a positive charge and a negative charge, they attract each other, like magnets do. Similarly, like charges (positive with positive, negative with negative) repel each other.
  • Measured in coulombs (C).
  • Smaller charges are often given in microcoulombs (\(\mu C\)), where 1 \(\mu C = 10^{-6} C\).
Charge plays a critical role in the creation of electric fields. These fields are responsible for the forces exerted by and on charged objects.
Studying how charges interact is essential to understanding capacitors, which store and release these charges.
Potential Difference
Potential difference, also known as voltage, is the work required to move a charge from one point to another in an electric field.
It's measured in volts (V) and represents the energy difference per charge between two points.
  • A high potential difference means lots of energy is available for moving charges.
  • A low potential difference means less available energy.
Think of potential difference like the pressure in a water hose. Higher pressure is similar to higher voltage, pushing charges more forcefully.
In a capacitor, the potential difference exists between its two plates, resulting in the storage of electric energy.
Capacitor
A capacitor is a device that stores electric charge and energy in an electric field. It consists of two conductors separated by an insulator.
These conductors are often referred to as plates. When connected to a power source, one plate collects positive charge while the other collects an equal negative charge.
  • The insulator prevents charges from moving between plates.
  • Capacitors are widely used in electronic circuits for energy storage.
Capacitors release stored energy quickly, making them ideal for things like camera flashes and tuning radios.
The most important property of a capacitor is its capacitance, determined by the size and distance of its plates and the nature of the insulating material.
Capacitance Calculation
Capacitance (\(C\)) is a measure of a capacitor's ability to store charge per unit voltage. It is expressed in farads (F), which can also be microfarads (\(\mu F\) or \(10^{-6} F\)).
The formula to calculate capacitance is:
\[C = \frac{Q}{V}\]
where \(Q\) is the charge in coulombs and \(V\) is the potential difference in volts.
  • To find capacitance, know the charge and voltage values.
  • Rearranging the formula helps find unknown values such as charge or potential difference.
In practical calculations, it's crucial to convert units when necessary (e.g., \(\mu C\) to C or \(\mu F\) to F).
This ensures correct results, such as finding the capacitance in exercise solutions or determining how much charge a capacitor can store.

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Most popular questions from this chapter

A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 \(\mathrm{mm}\) . If the separation is decreased to \(1.15 \mathrm{mm},\) what is the energy stored (a) if the capacitor is disconnected from the potential source so the charge on the plates remains constant, and (b) if the capacitor remains connected to the potential source so the potential difference between the plates remains constant?

(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?

A parallel-plate capacitor having plates 6.0 \(\mathrm{cm}\) apart is connected across the terminals of a 12 \(\mathrm{V}\) battery. (a) Being as quantitative as you can, describe the location and shape of the equipotential surface that is at a potential of \(+6.0 \mathrm{V}\) relative to the potential of the negative plate. Avoid the edges of the plates. (b) Do the same for the equipotential surface that is at \(+2.0 \mathrm{V}\) relative to the negative plate. (c) What is the potential gradient between the plates?

You are working on an electronics project requiring a variety of capacitors, but have only a large supply of 100 nF capacitors available. Show how you can connect these capacitors to produce each of the following equivalent capacitances: (a) \(50 \mathrm{nF},\) (b) \(450 \mathrm{nF},(\mathrm{c}) 25 \mathrm{nF},\) (d) 75 \(\mathrm{nF.}\)

A particle with a charge of \(+4.20 \mathrm{nC}\) is in a uniform electric field \(\vec{\boldsymbol{E}}\) directed to the left. It is released from rest and moves to the left; after it has moved \(6.00 \mathrm{cm},\) its kinetic energy is found to be \(+1.50 \times 10^{-6} \mathrm{J}\) (a) What work was done by the electric force? (b) What is the potential of the starting point with respect to the endpoint? (c) What is the magnitude of \(\vec{E} ?\)
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