How to find the energy of capacitance?

[Coderhk]

Homework Statement

Find the energy in the 2uf capacitor in the diagram attached in uploads.

Homework Equations

E=(1/2)CV^2
E=QV
C=Q/V
Capacitor in series : 1/(Equivalent Capacitor) =Σ1/(Ci)*
Capacitor in parallel: Equivalent Capacitor =∑Ci*
*i is the index of summation

The Attempt at a Solution

Attached in uploads, but here's a summary of what I did.
First we find the equivalent capacitance of 2.55micro farads
using Q=CV I get net charge is 1.533E-5 Coulombs
Knowing voltage drop of 1uf, find charge
Then the charge on 1.55uf is total charge minus charge on 1uf.
Afterwards I revert back to second diagram and all capacitor in series have same charge.
Find V at 2uf
Plug in E= QV
Done.[/B]

Textbook answer is 2.2E-7 J

 

[gneill]

I don't understand the number you've used for the voltage in your final calculation. Where did it come from?

Also, the answer you say your textbook gives looks to me to be incorrect (the power of ten used).

 

[Coderhk]

I don't understand the number you've used for the voltage in your final calculation. Where did it come from?
View attachment 219370

Also, the answer you say your textbook gives looks to me to be incorrect (the power of ten used).

Oops that number was suppose to be 4.665V. The 4.665 is from the line right before the last line.

 

[gneill]

Oops that number was suppose to be 4.665V. The 4.665 is from the line right before the last line.

Okay, you might want to run the numbers (for that final calculation) through your calculator again though.

 

[Coderhk]

Okay, you might want to run the numbers (for that final calculation) through your calculator again though.

I got 2.17E-5. After significant figures I get 2.2E-5. I guess the book just has the wrong exponent. Thanks for you help.

 

[gneill]

You're welcome.

I should mention that you did a lot of extra work to get to the solution, finding the total equivalent capacitance, the total charge, then removing the charge for the 1 μF capacitor, then finding the voltage on the 2 μF capacitor, then finally calculating its energy.

Since the series connection of C₂ and the pair C₃||C₄ is a branch connected across the 6.0 V battery, you could have ignored C₁ entirely for this problem. Nothing C₁ does can affect that potential across the branch.

As an alternative approach, you might have used the potential divider formula for series capacitors to go directly to the voltage across C₂, then applied your energy formula.