Energy Stored in a Charged Capacitor

In summary, the problem is asking for the energy stored in a capacitor connected to a 12.0-V battery with 54.0 μC of charge, using the formula 1/2CV^2. The attempt at a solution involves finding Q first using C=Q/V, then plugging into 1/2QV, but the answer does not match the back of the book. The correct answer is 3.24 x 10^-4 J, not 3.88 x 10^-4 J.
  • #1
shadyy
2
0

Homework Statement



A 12.0-V battery is connected to a capacitor, resulting in 54.0 μC of charge stored on the capacitor. How much energy is stored in the capacitor?


Homework Equations



1/2CV^2

The Attempt at a Solution



At first this problem looks to be easy for me but somehow I get a complete different answer than the back of the book.

I plugged in all the variables and solved and got 3.88 x 10^-4 J

1/2(5.4x10^-6)(12)^2

The answer I get in the back of the book is 3.24 x 10^-4 J

Can someone please explain to me what I am doing wrong?
 
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  • #2
shadyy said:

Homework Statement



A 12.0-V battery is connected to a capacitor, resulting in 54.0 μC of charge stored on the capacitor. How much energy is stored in the capacitor?


Homework Equations



1/2CV^2

The Attempt at a Solution



At first this problem looks to be easy for me but somehow I get a complete different answer than the back of the book.

I plugged in all the variables and solved and got 3.88 x 10^-6 J

1/2(5.4x10^-6)(12)^2

The answer I get in the back of the book is 3.24 x 10^-4 J

Can someone please explain to me what I am doing wrong?

Try finding Q first using C=Q/V, then try 1/2QV. This will either confirm that you are right and the book is wong, or that the book is right and that you are wrong.
 
  • #3
Thanks for the quick reply. I tried that and I got a completely different answer. I don't think the book is wrong. I think I'm doing something wrong. I'll try to figure it out I guess.
 
  • #4
Energy stored = 0.5QV
You know Q (charge) and you know V
 
  • #5




It seems like you have correctly used the formula for calculating the energy stored in a capacitor (1/2CV^2). However, there may be a discrepancy in the value of the capacitance used in your calculation. It is possible that the value of 5.4x10^-6 is not the actual capacitance of the capacitor in question, leading to a different answer than the one provided in the book. I would recommend double-checking the given values or consulting with your instructor for clarification. Additionally, it is always good practice to include units in your calculations to ensure accuracy.
 

1. What is a charged capacitor?

A charged capacitor is an electronic component that stores electrical energy in the form of an electric charge. It is made up of two conductive plates separated by a dielectric material.

2. How is energy stored in a charged capacitor?

When a capacitor is connected to a power source, such as a battery, it becomes charged with electrons. The electrons collect on one plate, creating a negative charge, while the other plate becomes positively charged. This separation of charges creates an electric field, which stores the energy in the capacitor.

3. What factors affect the amount of energy stored in a charged capacitor?

The amount of energy stored in a charged capacitor is affected by the capacitance, voltage, and geometry of the capacitor. A higher capacitance, voltage, and surface area of the plates will result in a greater energy storage capacity.

4. How is the energy stored in a charged capacitor calculated?

The energy stored in a charged capacitor can be calculated using the formula E = 1/2CV^2, where E is the energy in joules, C is the capacitance in farads, and V is the voltage in volts.

5. How is the energy released from a charged capacitor?

The energy stored in a charged capacitor can be released by connecting it to a circuit. As the electrons flow from one plate to the other, the capacitor discharges and releases its stored energy in the form of an electric current.

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