A dielectric-filled parallel-plate capacitor has plate area = 25.0 , plate separ

Question

A dielectric-filled parallel-plate capacitor has plate area = 25.0 , plate separation = 10.0 and dielectric constant = 4.00. The capacitor is connected to a battery that creates a constant voltage = 12.5 . Throughout the problem, use = 8.85×10-12 .

1- Find the energy of the dielectric-filled capacitor.

2- The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy of the capacitor at the moment when the capacitor is half-filled with the dielectric.

Answers in Joules.

Explanation / Answer

Inserting the dielectric increases the capacitance to

C' = C = oA/d

The problem is you didn't include the units in your problem, so I can't actually solve it properly. I'm just going to guess as to what the units are, and if the units are wrong, just change them and solve it yourself, or comment with the proper units and I can solve it for you in the comments.

C' = (4)(8.85*10^-12 F/m)(25 cm^2 * 10^-4 m^2/cm^2)/(10cm * 10^-2 m/ cm) = 8.85*10^-13 F

1- If the capacitor has a voltage across it of V = 12V, then

U = (1/2)CV^2 = (1/2)(8.85*10^-F)(12V)^2 = 6.37*10^-11 J

2- If you remove half of the dielectric, you now have essentially two capacitors with two seperate energies. The first capacitor has no dielectric, but has half the area, so it has a capacitance of

C1 = (8.85*10^-12 F/m)(12.5*10^-4 m^2)/(0.1m) = 1.11*10^-13 F

The other capacitor has the dielectric in it and has half the area also, so

C2 = (4)(8.85*10^-12 F/m)(12.5*10^-4 m^2)/(0.1m) = 4.43*10^-13 F

So, the total energy stored is the sum of the two energies produced by the two capacitors (remember, this is just a way of rationilizing the answer, there arent actually two capacitors, just one seperated into two regions),

U = U1 + U2 = (1/2)C1V^2 + (1/2)C2V^2 = (1/2)(C1+C2)V^2

= (1/2)(4.43*10^-13 F + 1.11*10^-13 F)(12V)^2 = 3.99*10^-11 J

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