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Chapter 20: Problem 93

Two capacitors are connected in parallel across the terminals of a battery. One has a capacitance of \(2.0 \mu \mathrm{F}\) and the other a capacitance of \(4.0 \mu \mathrm{F} .\) These two capacitors together store \(5.4 \times 10^{-5} \mathrm{C}\) of charge. What is the voltage of the battery?

Short Answer

Expert verified
The voltage of the battery is 9.0V.

Step by step solution

01

Identify Capacitance in Parallel

When capacitors are connected in parallel, their capacitances add up. Therefore, the total capacitance \( C_t \) is the sum of the individual capacitances. \[ C_t = C_1 + C_2 = 2.0 \mu F + 4.0 \mu F = 6.0 \mu F. \]
02

Understand Charge Stored

The total charge \( Q \) stored in capacitors connected in parallel is equal to the charge from the battery. Given that the total charge is \( 5.4 \times 10^{-5} \mathrm{C} \), we can proceed to find the voltage.
03

Calculate Voltage Using the Formula

To find the voltage \( V \) across the capacitors, use the formula \( Q = C_t \times V \). Substituting in the values, we get: \[ V = \frac{Q}{C_t} = \frac{5.4 \times 10^{-5} \mathrm{C}}{6.0 \times 10^{-6} \mathrm{F}} = 9.0 \mathrm{V}. \]

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Key Concepts

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

Understanding Capacitance
Capacitance is a key concept when working with capacitors. It refers to a capacitor's ability to store electric charge per unit of voltage. In simpler terms, capacitance tells us how much electric charge a capacitor can hold. The unit of capacitance is the farad (F), and because capacitors often have relatively small capacitance values, microfarads (\( \mu \F \)) are commonly used.
  • Capacitors in Parallel: When capacitors are connected in parallel, their total capacitance is the sum of their individual capacitances. In this setup, the combined capacitance increases, allowing the circuit to store more charge.
  • Calculation Example: For two capacitors with capacitances of \( 2.0 \mu F \) and \( 4.0 \mu F \), their total capacitance will be \( C_t = 2.0 \mu F + 4.0 \mu F = 6.0 \mu F \).
Exploring Electric Charge
Electric charge is fundamental to understanding how capacitors work. It is the property of matter that causes it to experience a force when near other electrically charged matter. The electric charge is typically measured in coulombs (C). In the context of capacitors, the charge they store is directly related to the voltage applied across them.
  • Charge Storage: When a voltage is applied to a capacitor, an electric field is created, causing positive charge to accumulate on one plate and negative charge on the other.
  • Role of Total Charge: The total charge stored in capacitors arranged in parallel is simply the sum of the charges on all capacitors, constrained by the shared voltage across them. For the given example, the capacitors together store \( 5.4 \times 10^{-5} \text{C}\).
Voltage Calculation in Circuits
Calculating voltage in circuits, especially ones including capacitors, involves understanding the relationship between charge, capacitance, and voltage. This is typically expressed in the formula: \[ Q = C \times V \]Where:- \( Q \): Charge in coulombs- \( C \): Capacitance in farads- \( V \): Voltage in volts
  • Using the Formula: To find the voltage when you know the charge and capacitance, simply rearrange the formula to \( V = \frac{Q}{C} \).
  • Worked Example: In our example, with \( Q = 5.4 \times 10^{-5} \mathrm{C} \) and \( C_t = 6.0 \mu F \) or \( 6.0 \times 10^{-6} \mathrm{F} \), the voltage is \( V = \frac{5.4 \times 10^{-5}}{6.0 \times 10^{-6}} = 9.0 \mathrm{V} \).

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Most popular questions from this chapter

The heating element in an iron has a resistance of \(24 \Omega .\) The iron is plugged into a \(120-\mathrm{V}\) outlet. What is the power delivered to the iron?

Multiple-Concept Example 9 discusses the physics principles used in this problem. Three resistors, \(2.0,4.0,\) and \(6.0 \Omega,\) are connected in series across a \(24-\mathrm{V}\) battery. Find the power delivered to each resistor

The temperature coefficient of resistivity for the metal gold is 0.0034 \(\left(\mathrm{C}^{\circ}\right)^{-1},\) and for tungsten it is \(0.0045\left(\mathrm{C}^{\circ}\right)^{-1} .\) The resistance of a gold wire increases by \(7.0 \%\) due to an increase in temperature. For the same increase in temperature, what is the percentage increase in the resistance of a tungsten wire?

A \(60.0-\Omega\) resistor is connected in parallel with a \(120.0-\Omega\) resistor This parallel group is connected in series with a \(20.0-\Omega\) resistor. The total combination is connected across a \(15.0-\mathrm{V}\) battery. Find (a) the current and (b) the power delivered to the \(120.0-\Omega\) resistor.

Three parallel plate capacitors are connected in series. These capacitors have identical geometries. However, they are filled with three different materials. The dielectric constants of these materials are \(3.30,5.40,\) and \(6.70 .\) It is desired to replace this series combination with a single parallel plate capacitor. Assuming that this single capacitor has the same geometry as each of the other three capacitors, determine the dielectric constant of the material with which it is filled.
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