An 8900-pF capacitor holds plus and minus charges of 1.85×10−7 C . Part A What is the voltage across the capacitor?
The Correct Answer and Explanation is :
The correct answer is: 20.8 volts
To find the voltage across a capacitor, we can use the relationship given by the formula:
[
V = \frac{Q}{C}
]
where:
- ( V ) is the voltage across the capacitor,
- ( Q ) is the charge stored in the capacitor, and
- ( C ) is the capacitance of the capacitor.
In this problem:
- The capacitance ( C ) is given as 8900 pF (picoFarads). We need to convert this to Farads for our calculations:
[
C = 8900 \, \text{pF} = 8900 \times 10^{-12} \, \text{F} = 8.9 \times 10^{-9} \, \text{F}
]
- The charge ( Q ) is given as ( 1.85 \times 10^{-7} \, \text{C} ). This is the magnitude of the charge stored, regardless of the sign, as capacitors hold positive and negative charges equally on their plates.
Now, we can substitute these values into the formula to find the voltage ( V ):
[
V = \frac{1.85 \times 10^{-7} \, \text{C}}{8.9 \times 10^{-9} \, \text{F}}
]
Calculating this gives:
[
V \approx \frac{1.85 \times 10^{-7}}{8.9 \times 10^{-9}} \approx 20.8 \, \text{V}
]
So, the voltage across the capacitor is approximately 20.8 volts.
Explanation:
The capacitance of a capacitor indicates how much charge it can store per unit of voltage. The formula ( V = \frac{Q}{C} ) allows us to determine the voltage across a capacitor when the amount of charge and the capacitance are known. In this case, we used the values for charge and capacitance to calculate the voltage.
A capacitor stores electrical energy in an electric field created between its plates when charged. The greater the capacitance, the more charge it can store at a given voltage, which is fundamental in many electronic applications. Understanding the relationship between charge, capacitance, and voltage is crucial in fields such as electronics and electrical engineering, where capacitors are commonly used for energy storage, filtering, and timing applications.