We appreciate your visit to The energy stored in a parallel plate capacitor is 3 J What is the capacitance of the capacitor if the potential difference between the plates. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
Sure, let's solve the given problem step-by-step.
Question:
The energy stored in a parallel plate capacitor is 3 Joules. What is the capacitance of the capacitor if the potential difference between the plates is 220 Volts?
Given data:
1. Energy stored (E) = 3 Joules
2. Potential difference (V) = 220 Volts
To find: Capacitance (C) in microfarads ([tex]\( \mu F \)[/tex])
Step-by-step solution:
1. Write down the formula for the energy stored in a capacitor:
[tex]\[
E = \frac{1}{2} C V^2
\][/tex]
2. Rearrange the formula to solve for capacitance (C):
[tex]\[
C = \frac{2E}{V^2}
\][/tex]
3. Substitute the given values into the formula:
[tex]\[
C = \frac{2 \times 3}{220^2}
\][/tex]
4. Calculate the capacitance in Farads:
[tex]\[
C = \frac{6}{48400} \approx 0.00012396694214876034 \, \text{Farads}
\][/tex]
5. Convert the capacitance from Farads to microfarads:
[tex]\[
1 \, \text{Farad} = 10^6 \, \mu \text{Farads}
\][/tex]
[tex]\[
C \approx 0.00012396694214876034 \, \text{Farads} \times 10^6 \approx 123.96694214876034 \, \mu \text{Farads}
\][/tex]
Result:
The capacitance of the capacitor is approximately [tex]\( 124 \, \mu F \)[/tex].
So, the correct answer is:
A. [tex]\( 124 \, \mu \text{F} \)[/tex]
Question:
The energy stored in a parallel plate capacitor is 3 Joules. What is the capacitance of the capacitor if the potential difference between the plates is 220 Volts?
Given data:
1. Energy stored (E) = 3 Joules
2. Potential difference (V) = 220 Volts
To find: Capacitance (C) in microfarads ([tex]\( \mu F \)[/tex])
Step-by-step solution:
1. Write down the formula for the energy stored in a capacitor:
[tex]\[
E = \frac{1}{2} C V^2
\][/tex]
2. Rearrange the formula to solve for capacitance (C):
[tex]\[
C = \frac{2E}{V^2}
\][/tex]
3. Substitute the given values into the formula:
[tex]\[
C = \frac{2 \times 3}{220^2}
\][/tex]
4. Calculate the capacitance in Farads:
[tex]\[
C = \frac{6}{48400} \approx 0.00012396694214876034 \, \text{Farads}
\][/tex]
5. Convert the capacitance from Farads to microfarads:
[tex]\[
1 \, \text{Farad} = 10^6 \, \mu \text{Farads}
\][/tex]
[tex]\[
C \approx 0.00012396694214876034 \, \text{Farads} \times 10^6 \approx 123.96694214876034 \, \mu \text{Farads}
\][/tex]
Result:
The capacitance of the capacitor is approximately [tex]\( 124 \, \mu F \)[/tex].
So, the correct answer is:
A. [tex]\( 124 \, \mu \text{F} \)[/tex]
Thanks for taking the time to read The energy stored in a parallel plate capacitor is 3 J What is the capacitance of the capacitor if the potential difference between the plates. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!
- Why do Businesses Exist Why does Starbucks Exist What Service does Starbucks Provide Really what is their product.
- The pattern of numbers below is an arithmetic sequence tex 14 24 34 44 54 ldots tex Which statement describes the recursive function used to..
- Morgan felt the need to streamline Edison Electric What changes did Morgan make.
Rewritten by : Barada