Energy of a Capacitor System with partially inserted Dielectric

  • Thread starter saad87
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Homework Statement


Picture2-3.jpg


The above is a capacitor system. A dielectric is inserted into capacitor C3 and I need to find the total energy of the system. The dielectric is not fully in, only partially (lets say a length x).


The Attempt at a Solution



I first found the capacitance of C3 with the dielectric partially inserted. I know that dielectric makes a system of 2 parallel capacitors.

Hence

[tex]C3 = \frac{L\epsilon_{0}}{d} (x(\epsilon_{r}-1) + L)[/tex]

where L is the length and width of the parallel capacitor.

Total capacitance of the system = [tex]\frac{(C3 + C2)C1}{C3 + C2 + C1}[/tex]

To find the total energy, I use the formula [tex]1/2 \times \frac{Q^{2}}{C}[/tex]

which becomes:

[tex]1/2 Q^{2} \times \frac{C3 + C2 + C1}{(C3 + C2)C1}[/tex]

Does this seem correct? The answer give to me by the professor is:

[tex]U =1/2 \frac{Q^{2}}{C1} + \frac{Q^{2}}{C2 + C3}[/tex]

I'm just kinda whether I'm wrong or the prof is...

Please advise.
 

Answers and Replies

  • #2
tiny-tim
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Hi saad87! :wink:
[tex]1/2 Q^{2} \times \frac{C3 + C2 + C1}{(C3 + C2)C1}[/tex]

Does this seem correct? The answer give to me by the professor is:

[tex]U =1/2 \frac{Q^{2}}{C1} + \frac{Q^{2}}{C2 + C3}[/tex]

I'm just kinda whether I'm wrong or the prof is...

Please advise.
erm :redface: … they're the same! (assuming the professor's "1/2" is for both bits) :smile:
 
  • #3
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Really? Could please show me how? I really can't understand how they are equal!

Do we use Partial fractions to expand the fractions or am I way off?
 
  • #4
tiny-tim
Science Advisor
Homework Helper
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1/C1 = (C2 + C3)/(C2 + C3)C1

1/(C2 + C3) = C1/(C2 + C3)C1 :wink:

(generally, 1/a + 1/b = (a + b)/ab )
 

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