Multi-Plate variable Capacitor Questions.

In summary: I really appreciate it. :DIn summary, the conversation discussed the estimation of the required radius for each plate of a multi-plate variable capacitor with 4 pairs of plates, separated in air by 0.01mm, and requiring a capacitance range of 10 - 400 pF. The equations for capacitance, initial current, and time constant were also mentioned. The conversation concluded with the correct calculation of the radius for the plates, with a final value of 6.41mm.
  • #1
oxon88
176
1

Homework Statement



A multi-plate variable capacitor has 4 pairs of plates. The plates, when closed, are separated in air by 0.01mm. A capacitance range of 10 - 400 pF is required

a) estimate the required radius, R of each plate.

b) The capacitor is set to the maximum of 400 pF and is charged to 10v throught a 50KΩ resistor. determine: i) the initial value of current flowing. ii) the time constant for the circuit.
IMAG0096.jpg

Homework Equations



capacitance = εr ε0A(n-1) / d

Initial current = V/R

Time constant = R x C

The Attempt at a Solution



a) 400x10-12 = εr ε0A(8-1) / (1x10-5)

can anyone explain what these 2 values are? (εr & ε0) i think one is dialectric strength. which for air is 3MV m-1 ?

b)
i) 10V/50Kohm = 0.2mA

ii) (50x103) x (400x10-12) = 20 μs
 
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  • #2
oxon88 said:

The Attempt at a Solution



can anyone explain what these 2 values are? (εr & ε0) i think one is dialectric strength. which for air is 3MV m-1 ?

##\epsilon_o = 8.8542\times10^{-12}\;F m^{-1}## is the permittivity of vacuum (free space)

##\epsilon_r = 1.00058986## is the relative permittivity for air
 
  • #3
Ok thanks, so...3. The Attempt at a Solution

a) 400x10-12 = 1 x (8.8542x10-12)A(8-1) / (1x10-5)re-arranging for A.

A = C.d / ϵorA = (400x10-12)*(1x10-5) / (8.8542x10-12)

A = 0.451763x10-3 = 0.451763 mm2A= ∏r2

r = √(a/∏)

r = √(0.451763x10-3/∏) = 11.99mm

can anyone check if this is ok?b)
i) 10V/50Kohm = 0.2mA

ii) (50x103) x (400x10-12) = 20 μsdo these answers look plausible?
 
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  • #4
oxon88 said:
Ok thanks, so...


3. The Attempt at a Solution

a) 400x10-12 = 1 x (8.8542x10-12)A(8-1) / (1x10-5)


re-arranging for A.

A = C.d / ϵor
Oops. What happened to the "(8 - 1)" ?
A = (400x10-12)*(1x10-5) / (8.8542x10-12)

A = 0.451763x10-3 = 0.451763 mm2


A= ∏r2
Are the plate areas full circles?
r = √(a/∏)

r = √(0.451763x10-3/∏) = 11.99mm

can anyone check if this is ok?


b)
i) 10V/50Kohm = 0.2mA

ii) (50x103) x (400x10-12) = 20 μs
These last two look fine.
 
  • #5
oops forgot about that. got carried away! :Da) 400x10-12 = 1 x (8.8542x10-12)A(8-1) / (1x10-5)re-arranging for A.

A = C.d / ϵor(n-1)A = (400x10-12)*(1x10-5) / (8.8542x10-12)(7)

A =0.0645376x10-3 = 0.06454mm

A= ∏r2

r = √(a/∏)

r = √(0.0645376x10-3/∏) = 4.532mmthey are half circles

so if i multiply the Area by 2.

0.064545x10-3*2 =0.1290752x10-3

then work out the radius:

r = √(0.1290752x10-3/∏) = 6.4098mm
 
Last edited:
  • #6
oxon88 said:
oops forgot about that. got carried away! :D


a) 400x10-12 = 1 x (8.8542x10-12)A(8-1) / (1x10-5)


re-arranging for A.

A = C.d / ϵor(n-1)

A = (400x10-12)*(1x10-5) / (8.8542x10-12)(7)

A =0.0645376x10-3 = 0.06454mm
I think you have a unit conversion problem; The area should be three orders of magnitude larger if it's in mm2 (and it's mm2, not mm as you've written). Note that an area of 0.06 mm2 would be practically microscopic! That should be a tipoff that something's amiss :wink:
A= ∏r2
Again, are the plates full circles?
 
  • #7
ok. I'm not really sure where I'm going wrong here.



a)

C = (ϵor(n-1)) / d

C = 400x10-12 Farads
ϵo = 8.8542×10−12Fm−1
ϵr = 1.00058986
d = 1x10-5 m


400x10-12 = 1 x (8.8542x10-12)A(8-1) / (1x10-5)


re-arranging for A.

A = C.d / ϵor(n-1) is equation ok?



A = (400x10-12)*(1x10-5) / 7(8.8542x10-12)

A =0.066896x10-3 mm2
 
  • #8
oxon88 said:
ok. I'm not really sure where I'm going wrong here.
a)

C = (ϵor(n-1)) / d

C = 400x10-12 Farads
ϵo = 8.8542×10−12Fm−1
ϵr = 1.00058986
d = 1x10-5 m400x10-12 = 1 x (8.8542x10-12)A(8-1) / (1x10-5)re-arranging for A.

A = C.d / ϵor(n-1) is equation ok?

Yes, that's fine.
A = (400x10-12)*(1x10-5) / 7(8.8542x10-12)

A = 6.45 x 10-5 m2

6.45 x 10-5 m2 x (106 mm2/m2) = 64.5 mm2
 
  • #9
ok thanks for that. I've never been great with unit conversions. :blushing:



so now do I multiply the calulated area by 2 because the plates are semicircles?

A = 64.5376mm2*2 = 129mm2


then work out the radius...

r = √(129/∏) = 6.41mm
 
  • #10
oxon88 said:
ok thanks for that. I've never been great with unit conversions. :blushing:
so now do I multiply the calulated area by 2 because the plates are semicircles?

A = 64.5376mm2*2 = 129mm2then work out the radius...

r = √(129/∏) = 6.41mm

That works fine. You could also have said that each plate is a semicircle so it has area

##A = \frac{1}{2}\pi r^2##

It works out to the same thing.
Looks like you're done :smile:
 
  • #11
many thanks for your help and patients.
 

Related to Multi-Plate variable Capacitor Questions.

What is a multi-plate variable capacitor?

A multi-plate variable capacitor is an electrical component that consists of two or more metal plates separated by a dielectric material. The distance between the plates can be varied, thus changing the capacitance of the capacitor.

How does a multi-plate variable capacitor work?

A multi-plate variable capacitor works by storing electric charge between its plates. The amount of charge that can be stored is directly proportional to the distance between the plates. By adjusting the distance, the capacitance of the capacitor can be changed.

What are the applications of a multi-plate variable capacitor?

Multi-plate variable capacitors are commonly used in electronic circuits for tuning radio or TV frequencies. They are also used in variable frequency oscillators, filters, and other electronic devices that require adjustable capacitance.

How is the capacitance of a multi-plate variable capacitor calculated?

The capacitance of a multi-plate variable capacitor can be calculated using the formula C = εA/d, where C is the capacitance, ε is the permittivity of the dielectric material, A is the area of the plates, and d is the distance between the plates.

What are the advantages of using a multi-plate variable capacitor?

One of the main advantages of using a multi-plate variable capacitor is its ability to adjust the capacitance to a specific value, making it useful in various electronic applications. It also has a high capacitance range and is relatively inexpensive compared to other types of variable capacitors.

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