Conceptual Question on Parallel Plate Capacitors

In summary, two capacitors with different charges and capacitances are connected in parallel, causing the charges to redistribute and resulting in the same voltage across both capacitors. The final voltage is reported as positive because the net charge cannot change.
  • #1
waters
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Homework Statement


Two capacitors, C1 and C2, are separately charged to 166 C and 348 C, respectively. They are then attached in parallel so that the + plate of one goes to the - plate of the other, and vice versa, as shown on the diagram below (notice how C2 was rotated 180 degrees before the capacitors were connected). C1 has a capacitance of 39.8 F, and C2 has a capacitance of 174 F. What is the final voltage across C1?

http://lon-capa.mines.edu/res/csm/csmphyslib/type56_capacitors/AttachedInParallel.jpg [Broken]

Homework Equations


V(C1+C2) = Q1 - Q2
V = (Q1-Q2)/(C1+C2)
V = .851 V

The Attempt at a Solution



My question is, why are the charges subtracted? What goes on in this circuit to make that happen? Why does this happen when capacitors are aligned in this manner? Also, why is the final voltage reported as positive, when it should be negative?
 
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  • #2
"Authorization Required" to your link.

ehild
 
  • #3
Sorry about that. It should be fine now.
 

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  • #4
When the capacitors are connected, the "circuit" is really two disconnected halves--a top half and a bottom half. It follows from conservation of charge that on the top half, with AD, you have +Q1 from C1 and -Q2 from C2, and the opposite charge on the BC half. This charge will redistribute itself over the plates so that the new voltages for the capacitors are the same (as is required for a parallel circuit), but the net charge Q1-Q2 cannot change.
 
  • #5


I would first clarify the terminology used in the question. The term "final voltage" can be misleading as there is no one "final" voltage in a parallel plate capacitor circuit. Rather, there is a potential difference or voltage between the two plates of each capacitor, which can be calculated using the formula V = Q/C, where Q is the charge on the capacitor and C is its capacitance.

To answer the question, the charges on each capacitor are subtracted because when capacitors are connected in parallel, the total charge on the circuit remains the same. This means that the charges on the two capacitors must add up to the initial total charge of 166 C + 348 C = 514 C. This is known as the principle of charge conservation.

In this particular circuit, the charges are subtracted because the plates of C1 and C2 are connected in opposite polarity. This means that the positive charge on one plate of C1 is attracted to the negative charge on one plate of C2, and vice versa. This results in a redistribution of charges between the two capacitors, with some charge moving from C1 to C2 and vice versa. This redistribution continues until the charges on both capacitors are equal, resulting in a final voltage of 0 V.

However, the question asks for the voltage across C1 only. To calculate this, we use the formula V = Q/C, where Q is the charge on C1 and C is its capacitance. We know that the charge on C1 is 166 C - x, where x is the amount of charge that has moved from C1 to C2. This means that the voltage across C1 can be calculated as V = (166 C - x)/39.8 F = 4.18 V - 0.851 V = 3.33 V.

As for the positive sign of the final voltage, this is simply a convention. In this case, it indicates that the voltage across C1 is in the same direction as the initial charge on C1. However, it is important to note that the actual potential difference between the two plates of C1 is negative, as the negative plate of C1 is now connected to the positive plate of C2. This is why it is important to use the correct terminology and understand the direction of the electric field in a circuit.

In summary, when capacitors are connected in parallel, the total charge remains the same and
 

What is a parallel plate capacitor?

A parallel plate capacitor is a device used to store electrical energy. It consists of two parallel conducting plates separated by an insulating material, known as a dielectric. The capacitor is able to store charge on its plates, creating an electric field between them.

How does a parallel plate capacitor work?

A parallel plate capacitor works by storing charge on its plates, creating an electric field between them. When a voltage is applied across the capacitor, electrons are attracted to one plate and repelled from the other, creating a potential difference. The amount of charge that can be stored on the plates is directly proportional to the voltage applied.

What factors affect the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor is affected by several factors, including the distance between the plates, the surface area of the plates, and the dielectric material between the plates. The capacitance is directly proportional to the surface area of the plates and inversely proportional to the distance between them. The type of dielectric material also plays a role, as some materials have a higher dielectric constant, leading to a higher capacitance.

What is the formula for calculating the capacitance of a parallel plate capacitor?

The formula for calculating the capacitance of a parallel plate capacitor is C=εA/d, where C is the capacitance, ε is the permittivity of the dielectric material, A is the surface area of the plates, and d is the distance between the plates. This formula assumes the plates are parallel, the electric field is uniform, and the dielectric material is isotropic.

What are some common applications of parallel plate capacitors?

Parallel plate capacitors have a variety of applications, including power factor correction, energy storage in electronic devices, and as filters in electrical circuits. They are also used in radio and television transmitters to tune frequencies and in sensors to measure pressure, humidity, and temperature. In addition, parallel plate capacitors are used in high voltage systems, such as power lines, to reduce the effects of power surges and improve overall efficiency.

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