Electric Current in Moving Plate Capacitor with Velocity 6 m/s and U=600 V

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In summary: The derivative of the product of two functions (CV) is not the product of the derivatives...In summary, the question is asking to find the electric current flowing in a circuit with a thin parallel plate capacitor connected to a voltage source of 600 V when one of the electrodes is pulled at a velocity of 6 m/s. The electrodes are square with a length of 10 cm and a distance of 1 mm between them. Using the equations C=ε0\frac{A}{d+vt}, C=\frac{q}{U}, and I=\frac{dq}{dt}, the current is not constant and depends on time. The complete equation for the current is i = d/dt (CV), not just i =
  • #1
Noirro
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Homework Statement


Thin parallel plate capacitor is connected to a U=600 V current source. Find the electric current flowing in the circuit if one of the electrodes is pulled at v=6 m/s velocity. The electrodes are square (l=10 cm), the distance between them is d=1 mm.

Homework Equations


C=ε0[itex]\frac{A}{d+vt}[/itex], C=[itex]\frac{q}{U}[/itex], I=[itex]\frac{dq}{dt}[/itex]

The Attempt at a Solution


I tried finding the current by first expressing q from those capacitor equations and plugging it into the dq/dt equation, but there is still time t left. I am not sure if the answer depends on time or is it possible to find some constant value of I?
 
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  • #2
Noirro said:

Homework Statement


Thin parallel plate capacitor is connected to a U=600 V current source. Find the electric current flowing in the circuit if one of the electrodes is pulled at v=6 m/s velocity. The electrodes are square (l=10 cm), the distance between them is d=1 mm.

Homework Equations


C=ε0[itex]\frac{A}{d+vt}[/itex], C=[itex]\frac{q}{U}[/itex], I=[itex]\frac{dq}{dt}[/itex]

The Attempt at a Solution


I tried finding the current by first expressing q from those capacitor equations and plugging it into the dq/dt equation, but there is still time t left. I am not sure if the answer depends on time or is it possible to find some constant value of I?

No, the current is not constant. Find the time dependence of the current, and the current at t=0.

ehild
 
  • #3
ehild said:
No, the current is not constant. Find the time dependence of the current, and the current at t=0.

ehild

OK thanks!
 
  • #4
Thin parallel plate capacitor is connected to a U=600 V current source

I assume you mean a voltage source.
 
  • #5
Hint: i = C dV/dt is not the complete story.

The complete story is i = d/dt (CV).
 

1. What is a capacitor?

A capacitor is a passive electrical component that stores energy in an electric field. It consists of two conductive plates separated by an insulating material, known as a dielectric.

2. How does a capacitor work?

When a voltage is applied to a capacitor, one plate becomes positively charged and the other becomes negatively charged, creating an electric field between them. This electric field stores the energy and allows the capacitor to hold a charge.

3. What factors affect the capacitance of a capacitor?

The capacitance of a capacitor is affected by the surface area of the plates, the distance between the plates, and the type of dielectric used. A larger surface area and smaller distance between plates result in a higher capacitance, while a higher dielectric constant leads to a higher capacitance.

4. How do you calculate the capacitance of a capacitor?

The capacitance of a capacitor can be calculated using the formula C = εA/d, where C is the capacitance in farads, ε is the permittivity of the dielectric material, A is the surface area of the plates in square meters, and d is the distance between the plates in meters.

5. What is a common application of capacitors?

Capacitors have a wide range of applications, including energy storage in electronic devices, power factor correction in electrical systems, and signal filtering in communication circuits. They are also used in timing circuits, motor starters, and power supplies.

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