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What voltage is required to store \(7.2 \times 10^{-5} \mathrm{C}\) of charge on the plates of a \(6.0-\mu \mathrm{F}\) capacitor?

Short Answer

Expert verified
12 volts.

Step by step solution

01

Understanding the Relationship Between Charge and Voltage

The relationship between charge, voltage, and capacitance is given by the formula \( Q = CV \), where \( Q \) is the charge in coulombs (C), \( C \) is the capacitance in farads (F), and \( V \) is the voltage in volts (V). To find the voltage \( V \), we can rearrange the equation to \( V = \frac{Q}{C} \).
02

Convert Capacitance to Farads

Capacitance given is \( 6.0 \) microfarads (\( \mu F \)), which needs to be converted to farads. Since \( 1 \mu F = 10^{-6} F \), \( 6.0 \mu F \) is \( 6.0 \times 10^{-6} F \).
03

Substitute the Known Values into the Equation

We need to substitute \( Q = 7.2 \times 10^{-5} C \) and \( C = 6.0 \times 10^{-6} F \) into the formula \( V = \frac{Q}{C} \). This gives us \( V = \frac{7.2 \times 10^{-5}}{6.0 \times 10^{-6}} \).
04

Calculate the Voltage

Perform the calculation: \[ V = \frac{7.2 \times 10^{-5}}{6.0 \times 10^{-6}} = 12 \text{ volts} \]. The voltage required is therefore 12 volts.

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Key Concepts

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

Charge-Voltage Relationship
The relationship between charge, voltage, and capacitance is fundamental in understanding how capacitors work. The formula \( Q = CV \) describes this relationship, where \( Q \) represents the charge in coulombs, \( C \) is the capacitance in farads, and \( V \) is the voltage in volts.To solve for voltage in these scenarios, we can rearrange the formula to \( V = \frac{Q}{C} \). This division tells us the voltage that needs to be applied across the capacitor plates to achieve a certain charge.- When given the charge \( Q \) and capacitance \( C \), use the rearranged formula to find \( V \).- It's important to ensure that the units for charge and capacitance are compatible before plugging them into the formula.
Capacitor Capacitance
Capacitor capacitance is a measure of the ability of a capacitor to store charge per unit voltage. It is quantified in farads (F). The higher the capacitance, the more charge a capacitor can store at a given voltage.Understanding capacitance allows us to determine the potential energy storage in electrical circuits. A key point to keep in mind is that even though farads might seem large, capacitors in typical electronic circuits often use smaller units like microfarads or picofarads.- 1 farad is quite large, often reserved for industrial applications.- Common capacitors in electronics often have capacitance in the range of microfarads (\( \mu F \)) or even lower.
Microfarads to Farads Conversion
To convert microfarads to farads, one simply needs to multiply the value in microfarads by \( 10^{-6} \). This conversion is crucial because it standardizes the units, allowing us to effectively use them in calculations like the charge-voltage relationship.For example, if you have a capacitance of \( 6.0 \) microfarads, it converts to \( 6.0 \times 10^{-6} \) farads.- The conversion formula is: \( 1 \mu F = 10^{-6} F \).- Always ensure that your capacitance is in farads when using the formula \( V = \frac{Q}{C} \), as this maintains consistency in your units and prevents calculation errors.

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Most popular questions from this chapter

A positive charge \(+q_{1}\) is located to the left of a negative charge \(-q_{2} .\) On a line passing through the two charges, there are two places where the total potential is zero. The first place is between the charges and is \(4.00 \mathrm{~cm}\) to the left of the negative charge. The second place is \(7.00 \mathrm{~cm}\) to the right of the negative charge. (a) What is the distance between the charges? (b) Find \(q_{1} / q_{2}\), the ratio of the magnitudes of the charges.

The electric potential energy stored in the capacitor of a defibrillator is \(73 \mathrm{~J}\), and the capacitance is \(120 \mu \mathrm{F}\). What is the potential difference that exists across the capacitor plates?

A positive point charge is surrounded by an equipotential surface \(A\), which has a radius of \(r_{A}\). A positive test charge moves from surface \(A\) to another equipotential surface \(B\), which has a radius of \(r_{B}\). In the process, the electric force does negative work. (a) Does the electric force acting on the test charge have the same or opposite direction as the displacement of the test charge? (b) Is \(r_{B}\) greater than or less than \(r_{A}\) ? Explain your answers. Problem The positive point charge is \(q=+7.2 \times 10^{-8} \mathrm{C}\), and the test charge is \(q_{0}=+4.5 \times 10^{-11} \mathrm{C}\). The work done as the test charge moves from surface \(A\) to surface \(B\) is \(W_{A B}=-8.1 \times 10^{-9} \mathrm{~J}\). The radius of surface \(A\) is \(r_{A}=1.8 \mathrm{~m}\). Find \(r_{B}\). Check to see that your answer is consistent with your answers to the Concept Questions.

The potential difference between the plates of a capacitor is \(175 \mathrm{~V}\). Midway between the plates, a proton and an electron are released. The electron is released from rest. The proton is projected perpendicularly toward the negative plate with an initial speed. The proton strikes the negative plate at the same instant that the electron strikes the positive plate. Ignore the attraction between the two particles, and find the initial speed of the proton.

Interactive LearningWare 19.2 at reviews the concepts pertinent to this problem. What is the potential difference between the plates of a 3.3-F capacitor that stores sufficient energy to operate a 75 -W light bulb for one minute?

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