Problem 38 - Physics

Question

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?

Step-by-Step Solution

Answer

12 volts.

Step 1: 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 \( V = \frac{Q}{C} \).

Step 2: 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 \( 6.0 \times 10^{-6} F \).

Step 3: 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 \( V = \frac{7.2 \times 10^{-5}}{6.0 \times 10^{-6}} \).

Step 4: 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.

Key Concepts

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.