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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 \(\mathrm{B}\) is 12 \(\mathrm{V}\) . Find the voltage across the plates of capacitor A.

Short Answer

Expert verified
The voltage across capacitor A is approximately 36.25 V.

Step by step solution

01

Understand the Relation Between Energy and Voltage for a Capacitor

The energy stored in a capacitor is given by the formula \( E = \frac{1}{2} C V^2 \), where \( E \) is the energy, \( C \) is the capacitance, and \( V \) is the voltage across the capacitor. We are given the energy for both capacitors and the voltage for capacitor B. We need to find the voltage for capacitor A using this formula.
02

Find the Capacitance of Capacitor B

Using the formula \( E = \frac{1}{2} C V^2 \), we can rearrange it to solve for the capacitance: \( C = \frac{2E}{V^2} \). For capacitor B, substitute \( E = 3.4 \times 10^{-4} \text{ J} \) and \( V = 12 \text{ V} \) into the formula: \[C_B = \frac{2 \times 3.4 \times 10^{-4}}{12^2}\]
03

Calculate Capacitance of Capacitor B

Compute the capacitance of capacitor B calculated from:\[C_B = \frac{2 \times 3.4 \times 10^{-4}}{12^2} = \frac{6.8 \times 10^{-4}}{144} = 4.72 \times 10^{-6} \text{ F}\]
04

Use Same Capacitance for Capacitor A

Since both capacitors are identical, the capacitance \( C_A \) of capacitor A is the same as that of capacitor B, which we found to be \( 4.72 \times 10^{-6} \text{ F} \).
05

Calculate the Voltage Across Capacitor A

We use the same energy formula for capacitor A. Rearrange it to solve for voltage: \( V = \sqrt{\frac{2E}{C}} \). Substitute \( E = 3.1 \times 10^{-3} \text{ J} \) and \( C = 4.72 \times 10^{-6} \text{ F} \) into the formula:\[V_A = \sqrt{\frac{2 \times 3.1 \times 10^{-3}}{4.72 \times 10^{-6}}}\]
06

Calculate Voltage for Capacitor A

Computing the voltage calculated:\[V_A = \sqrt{\frac{6.2 \times 10^{-3}}{4.72 \times 10^{-6}}} = \sqrt{1313.6} \approx 36.25 \text{ V}\]
07

Conclusion

The voltage across the plates of capacitor A is approximately \( 36.25 \text{ V} \).

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Key Concepts

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

Energy Storage in Capacitors
Capacitors are widely used in electrical circuits for storing energy. The energy stored in a capacitor is determined by the formula \( E = \frac{1}{2} C V^2 \). Here, \( E \) represents the energy in joules, \( C \) the capacitance in farads, and \( V \) the voltage across the capacitor in volts. This formula shows the direct relationship between energy storage and both capacitance and voltage:
  • An increase in capacitance or voltage results in more energy being stored.
  • Doubling the voltage, for instance, quadruples the energy stored due to the square dependence.
Capacitors can store significant amounts of energy for their size, making them indispensable in circuits needing quick energy release.
Voltage Across Capacitors
Voltage across a capacitor is crucial as it determines how much energy it can store and release. For two capacitors with identical capacitance but different stored energy, the difference in voltage is what makes them unique. In the exercise, capacitor B has a set voltage of 12 V, and it stores \( 3.4 \times 10^{-4} \) J of energy. To find the voltage for capacitor A, which stores more energy, the relation \( V = \sqrt{\frac{2E}{C}} \) is used:
  • With the same capacitance, capacitance \( C \) cancels out in calculations, allowing focus on energy values.
  • Larger stored energy in A means higher voltage compared to B, leading to the computed \( V_A \approx 36.25 \) V.
This demonstrates how a higher voltage challenges the same capacitive capacity to handle more energy.
Capacitance Calculation
Understanding capacitance is essential for working with circuits containing capacitors. Capacitance, denoted \( C \), indicates a capacitor's ability to store an electric charge per unit voltage and is crucial in determining how much energy a capacitor can hold. Using the formula \( C = \frac{2E}{V^2} \), capacitance can be calculated when energy and voltage are known.
  • For example, calculating the capacitance of capacitor B involves substituting its energy and voltage into this formula, resulting in a capacitance \( C_B = 4.72 \times 10^{-6} \) F.
  • With identical capacitors, these computed capacitance values can be directly applied to further calculations, as demonstrated with capacitor A.
Properly calculating capacitance allows precise control over circuit behavior, especially when deploying multiple capacitors with the same specifications.

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Most popular questions from this chapter

The membrane that surrounds a certain type of living cell has a surface area of \(5.0 \times 10^{-9} \mathrm{m}^{2}\) and a thickness of \(1.0 \times 10^{-8} \mathrm{m} .\) Assume that the membrane behaves like a parallel plate capacitor and has a dielectric constant of \(5.0 .\) (a) The potential on the outer surface of the membrane is \(+60.0 \mathrm{mV}\) greater than that on the inside surface. How much charge resides on the outer surface? (b) If the charge in part (a) is due to positive ions (charge \(+e ),\) how many such ions are present on the outer surface?

Multiple-Concept Example 4 to see the concepts that are pertinent here. In a television picture tube, electrons strike the screen after being accelerated from rest through a potential difference of 25000 \(\mathrm{V}\) . The speeds of the electrons are quite large, and for accurate calculations of the speeds, the effects of special relativity must be taken into account. Ignoring such effects, find the electron speed just before the electron strikes the screen.

Equipotential surface \(A\) has a potential of \(5650 \mathrm{V},\) while equipotential surface \(B\) has a potential of 7850 \(\mathrm{V}\) . A particle has a mass of \(5.00 \times 10^{-2} \mathrm{kg}\) and a charge of \(+4.00 \times 10^{-5} \mathrm{C}\) . The particle has a speed of 2.00 \(\mathrm{m} / \mathrm{s}\) on surface \(A\) . A nonconservative outside force is applied to the particle, and it moves to surface \(B\) , arriving there with a speed of 3.00 \(\mathrm{m} / \mathrm{s}\) . How much work is done by the outside force in moving the particle from \(A\) to \(B ?\)

Two particles each have a mass of \(6.0 \times 10^{-3}\) kg. One has a charge of \(+5.0 \times 10^{-6} \mathrm{C},\) and the other has a charge of \(-5.0 \times 10^{-6} \mathrm{C}\) . They are initially held at rest at a distance of 0.80 \(\mathrm{m}\) apart. Both are then released and accelerate toward each other. How fast is each particle moving when the separation between them is one-third its initial value?

Identical \(+1.8 \mu \mathrm{C}\) charges are fixed to adjacent corners of a square. What charge (magnitude and algebraic sign) should be fixed to one of the empty comers, so that the total electric potential at the remaining empty corner is 0 \(\mathrm{V} ?\)
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