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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) \(U=\frac{1}{2} C V^{2}\)

Short Answer

Expert verified
The energy can be expressed as \( U = \frac{Q^2}{2C} \) and \( U = \frac{1}{2} C V^2 \).

Step by step solution

01

Understanding the Given Expression

We start from the given expression for the energy stored in a capacitor, which is \( U = \frac{1}{2} Q V \). Here, \( Q \) is the charge on the capacitor, and \( V \) is the potential across the capacitor.
02

Using the Relation for Capacitance and Charge

The capacitance \( C \) is defined as \( C = \frac{Q}{V} \). Rearranging this formula gives \( Q = C V \). We'll use this in our calculations to derive the new expressions for energy.
03

Deriving Expression for \( U = \frac{Q^2}{2C} \)

We begin with the expression for \( Q \) in terms of \( C \) and \( V \): \( Q = C V \). Substitute \( V = \frac{Q}{C} \) into the original energy expression \( U = \frac{1}{2} Q V \). This gives \( U = \frac{1}{2} Q \left( \frac{Q}{C} \right) = \frac{Q^2}{2C} \).
04

Deriving Expression for \( U = \frac{1}{2} C V^2 \)

Substituting \( Q = C V \) into the original formula \( U = \frac{1}{2} Q V \), we have \( U = \frac{1}{2} (C V) V \). Simplify this to get \( U = \frac{1}{2} C V^2 \).
05

Verification of Both Forms

Both derived expressions, \( U = \frac{Q^2}{2C} \) and \( U = \frac{1}{2} C V^2 \), are consistent with the original form \( U = \frac{1}{2} Q V \) when using the relation \( C = \frac{Q}{V} \). This confirms that the energy stored in a capacitor can be represented in these forms.

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Key Concepts

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

energy storage
Capacitors are fascinating components in the world of electronics because they can store energy. Think of them like a small reservoir of electrical charge. When a capacitor is connected to a circuit, it can "hold" electric energy until it is needed. This stored energy can then be released to power devices or help with electrical signals.
That's why understanding how a capacitor stores energy is so important.
  • Capacitors are used in various everyday electronics like cameras and computers to store and release energy quickly.
  • The stored energy in a capacitor is key for devices that require a burst of energy.
  • This concept of energy storage makes capacitors versatile and essential components.
By storing energy, capacitors help stabilize voltage and power flow, ensuring devices operate smoothly and efficiently.
capacitance formula
The capacitance formula is the heart of understanding how capacitors work. Capacitance \(C\) refers to a capacitor's ability to store an electric charge. It's defined as the amount of charge \(Q\) stored per unit voltage \(V\) across the capacitor plates. The basic formula is \(C = \frac{Q}{V}\), where:
  • \(C\) is capacitance measured in farads (F).
  • \(Q\) represents charge measured in coulombs (C).
  • \(V\) is potential difference measured in volts (V).
This formula reveals a simple yet profound relationship: the greater the capacitance, the more charge can be stored at a given voltage. Furthermore, rearranging this formula can help us derive different expressions for the energy stored in a capacitor, such as \(U = \frac{1}{2} C V^2\) and \(U = \frac{Q^2}{2C}\). These expressions highlight how capacitance, charge, and potential are interlinked.
charge and potential
Charge and potential are fundamental to how capacitors function. The charge \(Q\) on a capacitor is the amount of electric charge stored on its plates. Potential, on the other hand, is the electrical pressure, or voltage \(V\), across the capacitor. These two aspects are closely related to the capacitance of the capacitor.
When a capacitor is connected to a power source, these charges accumulate, creating an electric field between the plates.
  • This electric field gives rise to the potential difference.
  • The stored charge equates to the capacitance multiplied by the potential difference, \(Q = C V\).
This relationship explains why altering the potential (voltage) changes the amount of charge a capacitor can store. The interplay of charge and potential allows capacitors to be flexible in handling various currents and voltages, making them integral components in electrical systems.
electrical potential energy
Electrical potential energy is the energy stored in a capacitor due to the charge on its plates. This energy can be calculated using different expressions based on the relationships among charge, capacitance, and voltage. One key formula is \(U = \frac{1}{2} Q V\), which represents the energy \(U\) as half of the product of charge and voltage. This formula highlights several factors:
  • High charge \(Q\) stores more energy.
  • High voltage \(V\) across the capacitor also means higher energy storage.
Other expressions like \(U = \frac{Q^2}{2C}\) and \(U = \frac{1}{2} C V^2\) can be derived from this. These demonstrate how energy storage depends on different parameters, such as charge and capacitance. Electrical potential energy is critical because it determines how much energy can be delivered to a circuit, influencing everything from small electronic devices to large power systems.

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Most popular questions from this chapter

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?

(a) \(A+5.00\) pC charge is located on a sheet of paper. (a) Draw to scale the curves where the equipotential surfaces due to these charges intersect the paper. Show only the surfaces that have a potential (relative to infinity) of \(1.00 \mathrm{V}, 2.00 \mathrm{V}\) \(3.00 \mathrm{V}, 4.00 \mathrm{V},\) and 5.00 \(\mathrm{V} .\) (b) The surfaces are separated equally in potential. Are they also separated equally in distance? (c) In words, describe the shape and orientation of the surfaces you just found.

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 metal sphere carrying an evenly distributed charge will have spherical equipotential surfaces surrounding it. Suppose the sphere's radius is 50.0 \(\mathrm{cm}\) and it carries a total charge of \(+1.50 \mu \mathrm{C}\) (a) Calculate the potential of the sphere's surface.(b) You want to draw equipotential surfaces at intervals of 500 \(\mathrm{V}\) outside the sphere's surface. Calculate the distance between the first and the second equipotential surfaces, and between the 20 \(\mathrm{th}\) and 21 \(\mathrm{st}\) equipotential surfaces. (c) What does the changing spacing of the surfaces tell you about the electric field?

Two oppositely charged identical insulating spheres, each 50.0 \(\mathrm{cm}\) in diameter and carrying a uniform charge of magnitude \(175 \mu \mathrm{C},\) are placed 1.00 \(\mathrm{m}\) apart center to center Fig. 18.53 ). (a) If a voltmeter is connected between the nearest points \((a\) and \(b)\) on their surfaces, what will it read? (b) Which point, \(a\) or \(b,\) is at the higher potential? How can you know this without any calculations?

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