Multi-Plate variable Capacitor Questions.

[oxon88]

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.

Homework Equations

capacitance = ε_r ε₀A(n-1) / d

Initial current = V/R

Time constant = R x C

The Attempt at a Solution

a) 400x10^-12 = ε_r ε₀A(8-1) / (1x10^-5)

can anyone explain what these 2 values are? (ε_r & 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

 

[gneill]

The Attempt at a Solution

can anyone explain what these 2 values are? (ε_r & 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

 

[oxon88]

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 / ϵ_o.ϵ_rA = (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?

 

[gneill]

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 / ϵ_o.ϵ_r

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.

 

[oxon88]

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 / ϵ_o.ϵ_r(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

 

[gneill]

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 / ϵ_o.ϵ_r(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

A= ∏r²

Again, are the plates full circles?

 

[oxon88]

ok. I'm not really sure where I'm going wrong here.



a)

C = (ϵ_o.ϵ<sub>r(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 / ϵo.ϵr(n-1) is equation ok?



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

A =0.066896x10-3 mm²</sub>

 

[gneill]

ok. I'm not really sure where I'm going wrong here.
a)

C = (ϵ_o.ϵ<sub>r(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 / ϵo.ϵr(n-1) is equation ok?</sub>

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

 

[oxon88]

ok thanks for that. I've never been great with unit conversions.



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

 

[gneill]

ok thanks for that. I've never been great with unit conversions.
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

 

[oxon88]

many thanks for your help and patients.