Parallel Plate Capacitor and Battery

[tristanm]

Homework Statement

Two parallel plates, each having area A = 3558 cm2 are connected to the terminals of a battery of voltage Vb = 6 V as shown. The plates are separated by a distance d = 0.54 cm.

1) What is Q the charge on the top plate? 3.5*10^-9 C
2) What is U, the energy stored in this capacitor? 1.05 * 10^-8 J
3) The battery is now disconnected from the plates and the separation of the plates is doubled ( = 1.08 cm). What is the energy stored in this new capacitor?
4) What is E, the magnitude of the electric field in the region between plates?
5) Compare V, the magnitude of the new potential difference across the plates to Vb, the voltage of the battery. V > Vb
6)Two uncharged parallel plates are now connected to the initial pair of plates as shown. How will the electric field, E, and potential difference across the plates, V, change, if at all? Both E and V will decrease?

Homework Equations

U = (1/2)*Epsilon-naught*E^2
Q=C*V

The Attempt at a Solution

So the problems I'm having are 3 and 4. For three, even if the battery is disconnected, the voltage will remain the same as well as the charges, and therefore the capacitance. Once the distance is doubled, the charge will remain the same, and the capacitance, meaning the voltage needs to be doubled to maintain C and Q at the same values. Therefore, the new U should equal (1/2)*Epsilon-naught*E^2 = (1/2)*Epsilon-naught*(V/d)^2

Plugging in V=12volts, d= 0.0054m and the standard 8.854*10^-12, I get 2.18*10^-7J which is incorrect.

As for question 5, wouldn't doing 12V/0.0054m give E? It isn't obviously, however I'm having trouble understanding why.

Thank you

 

[dauto]

You're using the wrong formula for U. There is a formula that relates U, Q, and V. Do you know it? How did you answer question 2?

For question 5. Didn't the distance double to 1.08 cm?

 

[tristanm]

You're using the wrong formula for U. There is a formula that relates U, Q, and V. Do you know it? How did you answer question 2?

For question 5. Didn't the distance double to 1.08 cm?

Thanks for the answer,

Yes it did double to 1.08 cm. It was a typo on my part.

In terms of question 2, I really have no idea. I played around with some equations and subbed them into each other and ended up with the correct answer. I can have a quick look at my notes when I get back home, however I can't recally the formula for U off the top of my head at the moment. Can I get a refresher?

 

[ehild]

Your formula for U is the energy density, energy per unit volume. You need the whole energy of the capacitor.

ehild

 

[HalfLostAndSearching]

for your help!

Dear student,

First of all, it is great to see that you are actively trying to solve these problems and seeking help when you encounter difficulties. Keep up the good work!

Now, let's address your questions. For number 3, you are correct in saying that the voltage will remain the same as the battery is disconnected. However, the charge on the plates will not necessarily remain the same. Remember that the charge on a capacitor is given by Q = CV, where C is the capacitance. In this case, the capacitance is not changing, but the voltage is. Therefore, the charge on the plates will also change. To find the new charge, you can use the fact that the capacitance is inversely proportional to the distance between the plates, so C' = C*d/d', where d' is the new distance. Using this, you can find the new charge on the plates and then use the equation U = (1/2)*QV to find the new energy stored in the capacitor.

For question 4, you are right in saying that the electric field can be found by dividing the voltage by the distance between the plates. However, the distance has now doubled, so the electric field will decrease by a factor of 2. This is because the electric field is inversely proportional to the distance between the plates, just like the capacitance. So for this problem, you will need to use the new distance of 1.08 cm to find the new electric field.

For question 5, your approach is correct. However, make sure you are using the correct units for the distance. The distance between the plates should be in meters, not centimeters. So the correct value for the electric field will be 12V/0.0108m = 1111.1 V/m.

For question 6, you are correct in saying that both the electric field and potential difference will decrease. This is because the newly added plates will increase the capacitance of the system, leading to a decrease in the electric field and potential difference.

I hope this helps clarify the problems for you. Keep up the good work!