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 is 220 V?

A. [tex]$124 \mu F$[/tex]
B. [tex]$660 \mu F$[/tex]
C. [tex]$75 \mu F$[/tex]
D. [tex]$124 mF$[/tex]

Given:
[tex]\[ E = \frac{1}{2} CV^2 \][/tex]
[tex]\[ 3 = \frac{1}{2} C (220)^2 \][/tex]

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]