# Multi-Plate variable Capacitor Questions.

1. Mar 27, 2012

### oxon88

1. The problem statement, all variables and given/known data

A multi-plate variable capacitor has 4 pairs of plates. The plates, when closed, are seperated 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.

2. Relevant equations

capacitance = εr ε0A(n-1) / d

Initial current = V/R

Time constant = R x C

3. 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

2. Mar 27, 2012

### Staff: Mentor

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

$\epsilon_r = 1.00058986$ is the relative permittivity for air

3. Mar 27, 2012

### 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 / ϵor

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

A = 0.451763x10-3 = 0.451763 mm2

A= ∏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 μs

Last edited: Mar 27, 2012
4. Mar 27, 2012

### Staff: Mentor

Oops. What happened to the "(8 - 1)" ?
Are the plate areas full circles?
These last two look fine.

5. Mar 27, 2012

### oxon88

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

A= ∏r2

r = √(a/∏)

r = √(0.0645376x10-3/∏) = 4.532mm

they are half circles

so if i multiply the Area by 2.

0.064545x10-3*2 =0.1290752x10-3

r = √(0.1290752x10-3/∏) = 6.4098mm

Last edited: Mar 27, 2012
6. Mar 27, 2012

### Staff: Mentor

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
Again, are the plates full circles?

7. Mar 27, 2012

### oxon88

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

a)

C = (ϵor(n-1)) / d

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. Mar 27, 2012

### Staff: Mentor

Yes, that's fine.
A = 6.45 x 10-5 m2

6.45 x 10-5 m2 x (106 mm2/m2) = 64.5 mm2

9. Mar 28, 2012

### 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

r = √(129/∏) = 6.41mm

10. Mar 28, 2012

### Staff: Mentor

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

11. Mar 28, 2012

### oxon88

many thanks for your help and patients.