Capacitor with sinusoidal voltage placed near a plate

In summary, the current output from an electrode is inversely proportional to the distance between the electrode and the torsional oscillator it is driving.
  • #1
unscientific
1,734
13

Homework Statement



An electrode with a varying voltage V = V0 + VACejωt is placed near a plate. The current output from the electrode is then measured. We want to find the relation between the current output and the amplitude of vibration. We can assume VAC is small relative to V0.


Homework Equations



Idea:

1. Constant charge in plate -> Constant force exerted -> separation changes -> capacitance changes
2. Apply sinusoidal voltage->phase change of force for every ∏ rad -> oscillation
3. these leads to changes in output current detected in electrode.

The Attempt at a Solution



V = V0 + VACe(jωt)
Q = CV
I = V*dC/dt + C*dV/dt
I = jCVACωejωt + V*dC/dt


C = ε0A/d
dC/dt = - (ε0A/d2) * d(d)/dt

I = jCVACωejωt + V*(ε0A/d2) * d(d)/dt

To find: d(d)/dt

F = Q2/(2ε0A)
m*dv/dt = Q2/(2ε0A)

v = Q2/(2mε0A) * t

therefore, the current is:

I = jCVACωejωt + Q3/(2ε0Am) * (1/d) * t

So the current depends both on time and inversely to displacement?
 
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  • #2
bumpp. My supervisor believes the current is proportional to the displacement though.
 
  • #3
It's application is on a torsional oscillator driven capacitively.
 
  • #4
If this is a driven system then there must be some sort of torsional spring restraining element between the plates pulling them apart. Because the plates are always drawn together by the applied voltage irrespective of its polarity. What is the torsional spring constant? What is d when the torsion torque is zero?

OR: are you sure this is not just a displacement sensor? In other words, is d not the independent variable? In which case force is not a consideration.
 
  • #5
rude man said:
If this is a driven system then there must be some sort of torsional spring restraining element between the plates pulling them apart. Because the plates are always drawn together by the applied voltage irrespective of its polarity. What is the torsional spring constant? What is d when the torsion torque is zero?

OR: are you sure this is not just a displacement sensor? In other words, is d not the independent variable? In which case force is not a consideration.

The plate is in fact a torsional oscillator, driven capacitively by the electrodes. So a sinusoidal voltage is applied to the electrode which attracts the oscillator at a certain frequency ω. The current depends on the capacitance, which in turn depends on the distance between the electrode and the oscillator. I want to find the relationship between the current observed and the distance between them.
 
  • #6
unscientific said:
I want to find the relationship between the current observed and the distance between them.

So why are you pursuing force on the plates? As I suspected, d is the independent variable. You already did a lot of analysis & came up with (incl. sign correction)
I = jCVACωejωt - V*(ε0A/d2) * d(d)/dt

I suggest replacing d with h to avoid confusion.

So that's I = jCVacωejωt - V*(ε0A/h2) * dh/dt

Now, replacing C with ε0A/h and V with V0 + Vacejwt you get

I = jwε0AVac/h ejwt - (V0 + Vacejwt0A (1/h2) dh/dt.

And that is as far as you can go without knowing what h(t) is.
Of course, replace each term with its real part to get the actual dependence of I on h.

Note that if dh/dt = 0 you get the standard relation between current and voltage of a fixed capacitor.
 
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  • #7
rude man said:
So why are you pursuing force on the plates? As I suspected, d is the independent variable. You already did a lot of analysis & came up with
I = jCVACωejωt + V*(ε0A/d2) * d(d)/dt

I suggest replacing d with h to avoid confusion.

So that's I = jCVacωejωt + V*(ε0A/h2) * dh/dt

Now, replacing C with ε0A/h and V with V0 + Vacejwt you get

I = jwε0A/h ejwt - (V0 + Vacejwt0A (1/h2) dh/dt.

And that is as far as you can go without knowing what h(t) is.
Of course, replace each term with its real part to get the actual dependence of I on h.

Note that if dh/dt = 0 you get the standard relation between current and voltage of a fixed capacitor.

So I thought. I'm not sure how he got it as linearly related to d; I'll have to look at his working when I go back to the lab and i'll post it here.
 
  • #8
unscientific said:
So I thought. I'm not sure how he got it as linearly related to d; I'll have to look at his working when I go back to the lab and i'll post it here.

Thanks. Only thing I can think of is if dh/dt is small, but even then I goes as 1/h, not h.

I suspect we don't have the complete picture. Could you post the question as originally posed? Thanks for posting the answer also, wish more OP's would do that.
 

Related to Capacitor with sinusoidal voltage placed near a plate

What is a capacitor?

A capacitor is an electronic component that stores electrical energy by accumulating an electrical charge on two conductive plates separated by an insulating material.

What is the purpose of a capacitor?

The main purpose of a capacitor is to store and release electrical energy in a circuit. It can also be used to filter out unwanted frequencies and stabilize voltage levels.

What happens when a sinusoidal voltage is applied to a capacitor?

When a sinusoidal voltage is applied to a capacitor, the capacitor will charge and discharge according to the frequency and amplitude of the voltage. This results in a sinusoidal current flowing through the capacitor.

How does a sinusoidal voltage affect a capacitor placed near a plate?

A sinusoidal voltage placed near a plate of a capacitor will cause the capacitor to charge and discharge at the same frequency as the applied voltage. This will result in a changing electric field between the plates, causing a changing electric flux and a sinusoidal current to flow through the capacitor.

What are the effects of placing a capacitor near a plate with a sinusoidal voltage?

Placing a capacitor near a plate with a sinusoidal voltage will result in a changing electric field and current flow through the capacitor. This can have various effects depending on the circuit, such as filtering out certain frequencies or storing energy for later use.

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