Capacitor Problem: Charge, Electric Displacement, and Current Calculations

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In summary, the conversation discusses the application of an electric potential of Vo*sin(wt) to a parallel plate capacitor of area S, separation d, and capacitance C. The questions revolve around finding the charge on the plates, the electric displacement current, and the current in the wire. The equations used are Q=CV, I=(epsilon)*d(phi)/dt, and i=dQ/dt. The solutions obtained are Q=CV=C*Vo*sin(wt) for part a, I=(epsilon)*w*cos(wt) for part b, and i=C*w*cos(wt) for part c. However, the ease of the solutions leads to a question about their accuracy. Additionally, the difference between displacement current and current
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TPDC130
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Homework Statement


Given an electric potential of Vo*sin(wt). This is applied to a parallel plate capacitor of area S, separation d and capacitance C. The questions are
a) What is charge Q on the plates?
b) Find the electric displacement current
c) Calculate the current in the wire.


Homework Equations


a) Q=CV
b) I=(epsilon)*d(phi)/dt
c) i=dQ/dt

The Attempt at a Solution



for part a, i got Q=CV=C*Vo*sin(wt)
for part b, i got I=(epsilon)*w*cos(wt)
for part c, i got i=C*w*cos(wt)


is this right, because i have a feeling that this is a little too easy of an answer.
why does the displacement current differ from the current in the wire?
 
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Rewrite that with [tex]\LaTeX[/tex] code,I think it`s easy as it can be.

It`s pain to read like that.
 

FAQ: Capacitor Problem: Charge, Electric Displacement, and Current Calculations

What is a capacitor?

A capacitor is an electronic component that stores electrical energy. It consists of two conductive plates separated by an insulating material, known as a dielectric. When a voltage is applied across the plates, an electric field is created, causing the plates to store electrical charge.

How do capacitors work?

Capacitors work by storing charge on their plates. When a voltage is applied, the electrons on one plate are repelled and move towards the other plate, creating a potential difference between the plates. The amount of charge that a capacitor can store depends on its capacitance, which is determined by the size of the plates and the distance between them.

What are the common uses of capacitors?

Capacitors have a wide range of applications in electronic circuits. Some common uses include filtering out unwanted signals, storing energy for power backup, and smoothing out voltage fluctuations. They are also used in timing circuits, audio equipment, and as part of power supplies.

What are the factors that affect the capacitance of a capacitor?

The capacitance of a capacitor depends on several factors, including the surface area of the plates, the distance between the plates, and the type of dielectric material used. Capacitance is also affected by the type of material used for the plates and the temperature of the capacitor.

How do I calculate the capacitance of a capacitor?

The capacitance of a capacitor can be calculated using the formula C = Q/V, where C is the capacitance in farads, Q is the charge stored on the plates in coulombs, and V is the voltage applied to the plates in volts. The capacitance can also be calculated by multiplying the permittivity of the dielectric by the area of the plates and dividing by the distance between them.

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