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Chapter 19: Problem 44

Two identical capacitors store different amounts of energy: capacitor A stores \(3.1 \times 10^{-3} \mathrm{J},\) and capacitor \(\mathrm{B}\) stores \(3.4 \times 10^{-4} \mathrm{J} .\) The voltage across the plates of capacitor B is 12 V. Find the voltage across the plates of capacitor A.

Short Answer

Expert verified
The voltage across capacitor A is approximately 36.14 V.

Step by step solution

01

Recall the Energy Formula for a Capacitor

The energy stored in a capacitor is given by the formula \( E = \frac{1}{2}CV^2 \), where \( E \) is the energy, \( C \) is the capacitance, and \( V \) is the voltage across the capacitor.
02

Find Capacitance of Capacitor B

Using the energy formula \( E = \frac{1}{2}CV^2 \), solve for capacitance \( C \). For capacitor B: \( E_B = 3.4 \times 10^{-4} \text{ J} \) and \( V_B = 12 \text{ V} \). Substitute to find \( C_B = \frac{2E_B}{V_B^2} \).Calculate:\[ C_B = \frac{2 \times 3.4 \times 10^{-4}}{12^2} \approx 4.72 \times 10^{-6} \text{ F} \].
03

Use Capacitance to Find Voltage for Capacitor A

Since the capacitors are identical, \( C_A = C_B = 4.72 \times 10^{-6} \text{ F} \). Use \( E = \frac{1}{2}CV^2 \) for capacitor A with \( E_A = 3.1 \times 10^{-3} \text{ J} \) to find \( V_A \).Rearrange to find:\[ V_A = \sqrt{\frac{2E_A}{C_A}} \].
04

Calculate Voltage Across Capacitor A

Substitute the values into the formula:\[ V_A = \sqrt{\frac{2 \times 3.1 \times 10^{-3}}{4.72 \times 10^{-6}}} \].Perform the calculation to find:\[ V_A \approx 36.14 \text{ V} \].

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Key Concepts

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

Energy Stored in a Capacitor
The energy stored in a capacitor is a fundamental concept in electronics, often used in circuits for applications like filtering and timing. Energy in a capacitor is expressed by the formula \( E = \frac{1}{2}CV^2 \). This equation shows that the energy stored depends on both the capacitance \( C \) and the voltage \( V \) across the capacitor's plates. In simpler terms:
  • \( C \) is how much charge a capacitor can hold per volt.
  • \( V \) is the electric potential difference across the plates.
Substituting these values into the formula lets you calculate the energy \( E \) in joules. Always ensure units are consistent, using farads for capacitance and volts for voltage. This energy can be converted back and forth, affecting other components in a circuit.
Capacitance
Capacitance itself is a measure of a capacitor's ability to hold an electric charge. Expressed in farads, capacitance is often quite small, in the microfarads (\( \mu F \)) or picofarads (\( pF \)), given that capacitors in practical use can't store large amounts of energy compared to batteries. Capacitance is determined by:
  • The surface area of the plates (larger area means more capacitance).
  • The distance between the plates (closer plates increase capacitance).
  • The dielectric material between the plates (better dielectrics increase capacitance).
Because the capacitors in the exercise were identical, their design parameters like plate area and separation were the same, leading them to have the same capacitance.
Voltage and Energy Relationship
Voltage and energy in a capacitor are intertwined, as evidenced by the formula \( E = \frac{1}{2}CV^2 \). As you increase the voltage across a capacitor, the energy stored rises quadratically. This means that even small increases in voltage can substantially increase energy. From the solution, you can observe:
  • Higher voltage in capacitor A (36.14 V) means more energy stored compared to capacitor B (12 V).
  • The nonlinear relationship means voltage changes have a more significant impact on energy than linear changes in capacitance.
Understanding this relationship helps when using capacitors in circuits that require precise energy management or when trying to store energy efficiently.
Identical Capacitors
In this context, identical capacitors refer to capacitors having the same capacitance value. Even though they might store different amounts of energy, as shown in the exercise, their ability to hold charge does not change unless physically altered. Key points to remember:
  • Identical capacitors store different energy amounts if voltage differs.
  • They are interchangeable in a circuit for similar tasks requiring the same capacitance.
  • Design parameters, e.g., plate size and dielectric, are the same in identical capacitors.
Thus, while capacitance stays the same, the voltage applied can vary to meet the circuit's needs, affecting how much energy each capacitor stores.

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Most popular questions from this chapter

A positive point charge \(\left(q=+7.2 \times 10^{-8} \mathrm{C}\right)\) is surrounded by an equipotential surface \(A,\) which has a radius of \(r_{A}=1.8 \mathrm{m} .\) A positive test charge \(\left(q_{0}=+4.5 \times 10^{-11} \mathrm{C}\right)\) moves from surface \(A\) to another equipotential surface \(B,\) which has a radius \(r_{B} .\) The work done as the test charge moves from surface \(A\) to surface \(B\) is \(W_{A B}=-8.1 \times 10^{-9}\) J. Find \(r_{B}\).

A parallel plate capacitor has a capacitance of \(7.0 \mu \mathrm{F}\) when filled with a dielectric. The area of each plate is \(1.5 \mathrm{m}^{2}\) and the separation between the plates is \(1.0 \times 10^{-5} \mathrm{m} .\) What is the dielectric constant of the dielectric?

An electric field has a constant value of \(4.0 \times 10^{3} \mathrm{V} / \mathrm{m}\) and is directed downward. The field is the same everywhere. The potential at a point \(P\) within this region is 155 V. Find the potential at the following points: (a) \(6.0 \times 10^{-3} \mathrm{m}\) directly above \(P,(\) (b) \(3.0 \times 10^{-3} \mathrm{m}\) directly below \(P\) (c) \(8.0 \times 10^{-3} \mathrm{m}\) directly to the right of \(P\)

Two identical point charges \(\left(+2.4 \times 10^{-9} \mathrm{C}\right)\) are fixed in place, separated by 0.50 \(\mathrm{m}\) (see the figure). Concepts: (i) The electric field is a vector and has a direction. At the midpoint, what are the directions of the individual electric-field contributions from \(q_{\mathrm{A}}\) and \(q_{\mathrm{B}} ?\) (ii) Is the magnitude of the net electric field at the midpoint greater than, less than, or equal to zero? (iii) Is the total electric potential at the midpoint positive, negative, or zero? (iv) Does the electric potential have a direction associated with it? Explain. Calculations: Find the electric field and the electric potential at the midpoint of the line between the charges \(q_{\mathrm{A}}\) and \(q_{\mathrm{B}}\).

Capacitor A and capacitor B both have the same voltage across their plates. However, the energy of capacitor A can melt \(m\) kilograms of ice at \(0^{\circ} \mathrm{C},\) while the energy of capacitor \(\mathrm{B}\) can boil away the same amount of water at \(100^{\circ} \mathrm{C} .\) The capacitance of capacitor \(\mathrm{A}\) is \(9.3 \mu \mathrm{F} .\) What is the capacitance of capacitor B?
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