Energy of a Capacitor System with partially inserted Dielectric

[saad87]

Homework Statement

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

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

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

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

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

which becomes:

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

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

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

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

Please advise.

 

[tiny-tim]

Hi saad87!

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

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

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

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

Please advise.

erm … they're the same! (assuming the professor's "1/2" is for both bits)

 

[saad87]

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?

 

[tiny-tim]

1/C₁ = (C₂ + C₃)/(C₂ + C₃)C₁

1/(C₂ + C₃) = C₁/(C₂ + C₃)C₁

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