College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![### Problem Statement
A 14.0-V battery is connected to a 4.20-µF capacitor. How much energy is stored in the capacitor?
\[ \boxed{\phantom{1}} \text{ J} \]
### Explanation
To solve this problem, you need to calculate the energy stored in the capacitor using the formula:
\[
E = \frac{1}{2} C V^2
\]
where:
- \( E \) is the energy stored in joules (J),
- \( C \) is the capacitance in farads (F),
- \( V \) is the voltage in volts (V).
### Given Values
- Voltage (V): 14.0 V
- Capacitance (C): 4.20 µF
### Solution Steps
1. Convert the capacitance from microfarads to farads:
\[
C = 4.20 \, \mu\text{F} = 4.20 \times 10^{-6} \, \text{F}
\]
2. Substitute the values into the formula:
\[
E = \frac{1}{2} \times 4.20 \times 10^{-6} \times (14.0)^2
\]
3. Calculate the energy stored:
\[
E = \frac{1}{2} \times 4.20 \times 10^{-6} \times 196
\]
\[
E = \frac{1}{2} \times 823.2 \times 10^{-6}
\]
\[
E = 411.6 \times 10^{-6}
\]
\[
E = 0.4116 \, \text{J}
\]
The energy stored in the capacitor is approximately 0.4116 joules.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9413ee1b-018f-4e4f-9457-fb88f0adac84%2F283b2e86-3d69-43fd-aebf-9ebfef84a846%2F0ehxxz_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
A 14.0-V battery is connected to a 4.20-µF capacitor. How much energy is stored in the capacitor?
\[ \boxed{\phantom{1}} \text{ J} \]
### Explanation
To solve this problem, you need to calculate the energy stored in the capacitor using the formula:
\[
E = \frac{1}{2} C V^2
\]
where:
- \( E \) is the energy stored in joules (J),
- \( C \) is the capacitance in farads (F),
- \( V \) is the voltage in volts (V).
### Given Values
- Voltage (V): 14.0 V
- Capacitance (C): 4.20 µF
### Solution Steps
1. Convert the capacitance from microfarads to farads:
\[
C = 4.20 \, \mu\text{F} = 4.20 \times 10^{-6} \, \text{F}
\]
2. Substitute the values into the formula:
\[
E = \frac{1}{2} \times 4.20 \times 10^{-6} \times (14.0)^2
\]
3. Calculate the energy stored:
\[
E = \frac{1}{2} \times 4.20 \times 10^{-6} \times 196
\]
\[
E = \frac{1}{2} \times 823.2 \times 10^{-6}
\]
\[
E = 411.6 \times 10^{-6}
\]
\[
E = 0.4116 \, \text{J}
\]
The energy stored in the capacitor is approximately 0.4116 joules.
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