Parallel Plate Capacitor Conundrum

In summary, the fraction of voltage across the air in a parallel plate capacitor with an air and dielectric gap can be calculated using the voltage divider formula and the dielectric constants of the two materials. The breakdown voltage of air can also be used to determine the voltage at which the air will begin to discharge.
  • #1
James11111
4
0

Homework Statement


Hi everyone, I'm new here at posting but have found this forum very useful in the past. I have hit a snag with a capacitor problem... any thoughts?

Assume you have a parallel plate capacitor separated by one cm of air and one cm of a dielectric with a dielectric constant of six. What fraction of the voltage between the two plates is across the air? Look up the breakdown strength of air and calculate the voltage when the air will begin to discharge.

ε = 6
ε0= 1.0059
d = 1 cm
Breakdown strength of air is 3MV/m

Homework Equations


V= I X

X = 1/jωC = d/(j ε ε0A)

V= I d / j ω A Ɛ Ɛ_0

The Attempt at a Solution



I started trying to add the capacitances of the two dielectics together in series, however, can I assume that the capacitance will divide according to their dielectric constants?
That is the voltage gradient will be 6 times stronger over the air compared to the dielectric?
 
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  • #2
James11111 said:

Homework Statement


Hi everyone, I'm new here at posting but have found this forum very useful in the past. I have hit a snag with a capacitor problem... any thoughts?

Assume you have a parallel plate capacitor separated by one cm of air and one cm of a dielectric with a dielectric constant of six. What fraction of the voltage between the two plates is across the air? Look up the breakdown strength of air and calculate the voltage when the air will begin to discharge.

ε = 6
ε0= 1.0059
d = 1 cm
Breakdown strength of air is 3MV/m

Homework Equations


V= I X

X = 1/jωC = d/(j ε ε0A)

V= I d / j ω A Ɛ Ɛ_0



The Attempt at a Solution



I started trying to add the capacitances of the two dielectics together in series, however, can I assume that the capacitance will divide according to their dielectric constants?
That is the voltage gradient will be 6 times stronger over the air compared to the dielectric?

Hello James11111, Welcome to Physics Forums.

What's the relationship between two capacitors of identical dimensions when one has no dielectric (air's dielectric constant is negligibly different from that of vacuum) and one that is filled with a dielectric? How does voltage divide across series capacitors?
 
  • #3
Yep figured the first bit.

It becomes a simple voltage divider circuit.

http://upload.wikimedia.org/wikipedia/en/f/f4/Impedance_Voltage_divider.png

k1=1.0006
k2=6.0

Vout = [Z2/(Z1+Z2)]*Vin

Zx = 1/j[itex]\omega[/itex]Cx - Sub in for each Z

Cx = kx * ε0*A/d - Sub in for each C

Vout = [k1/(k1+k2)]*Vin - After cancelling and simplifying.


The fraction of voltage over the dielectric component after substituting in k values is:
Vout = [1/(1+6)]*Vin=1/7*Vin

Which means that the fraction of voltage over the air must be 6/7.
 
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  • #4
Now to work out at what voltage the air will breakdown and discharge.

The breakdown strength of air is 3 MV/m, right?

This maybe too simplified but can I do the following?

distance between plates = 0.02 m

breakdown voltage of air over 0.02 m is 3M * 0.02 = 60,000 V

the fraction of 60,000V stressed across the air gap of the capacitor 6/7.

therefore, the air will begin to discharge when the voltage reaches:
VDischarge = (6/7)*60,000 = 51,428.6V
 
  • #5
James11111 said:
Now to work out at what voltage the air will breakdown and discharge.

The breakdown strength of air is 3 MV/m, right?

This maybe too simplified but can I do the following?

distance between plates = 0.02 m

breakdown voltage of air over 0.02 m is 3M * 0.02 = 60,000 V
Whoops. The air gap is not 2 cm wide -- half of that space is filled with dielectric. The 'air capacitor' gap is half that width. The dielectric surface serves as one plate of the air capacitor.
the fraction of 60,000V stressed across the air gap of the capacitor 6/7.

therefore, the air will begin to discharge when the voltage reaches:
VDischarge = (6/7)*60,000 = 51,428.6V
 
  • #6
Aha, Cheers!
 

FAQ: Parallel Plate Capacitor Conundrum

1. What is a parallel plate capacitor?

A parallel plate capacitor is a type of capacitor that consists of two parallel conductive plates separated by a non-conductive material, also known as a dielectric. It is used to store electrical energy by creating an electric field between the plates.

2. How does a parallel plate capacitor work?

When a voltage is applied to a parallel plate capacitor, an electric field is created between the plates. This causes a separation of positive and negative charges on each plate, resulting in an overall charge on the capacitor. The amount of charge stored on the capacitor is directly proportional to the applied voltage.

3. What is the equation for calculating the capacitance of a parallel plate capacitor?

The capacitance (C) of a parallel plate capacitor can be calculated using the equation C = ε0A/d, where ε0 is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.

4. How does the distance between the plates affect the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor is inversely proportional to the distance between the plates. This means that as the distance between the plates decreases, the capacitance increases, and vice versa. This is because a smaller distance allows for a stronger electric field and therefore more charge can be stored on the plates.

5. What are some practical applications of parallel plate capacitors?

Parallel plate capacitors have many practical applications, including in electronic devices such as computers and smartphones. They are also used in power factor correction, energy storage systems, and as sensors in various industries. Additionally, parallel plate capacitors are used in scientific research to study electric fields and their effects.

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