1. The problem statement, all variables and given/known data
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?
2. Relevant equations
U = (1/2)*Epsilon-naught*E^2
Q=C*V
3. 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
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?
2. Relevant equations
U = (1/2)*Epsilon-naught*E^2
Q=C*V
3. 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
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