# What must the radius of the plates in the designed capacitor (physics)?

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A 5.00 -pF parallel-plate air-filled capacitor with circular plates is to be used in a circuit in which it will be subjected to potentials of up to 1.00*10^2V. The electric field between the plates is to be no greater than 1.00*10^4 N/C. As a budding electrical engineer for Live-Wire Electronics, your task is to design the capacitor by finding what its physical dimensions and separation must be.

1. What must the radius of the plates in the designed capacitor be?

R=1.00*10^-2 cm, <— wrong, idk why….

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2.What must the separation of the plates in the designed capacitor be?

d= 4.2 cm <—- wrong,, idk why…

3.Find the maximum charge these plates can hold.

Q= 500 C <— wrong, ……

• 2. d = V/E = 0.01 m

1. C = e0*area/d ==> area = Cd/e0 = 0.005647 m^2

r = sqrt(area/pi) = 4.24E-2 m

3. Q = VC = 5E-10 C

Since you have to solve for answer 2 (value of d) first, it looks like you just swapped answers 1 and 2.

In 3 you seem to have ignored the “p” in “pF”. That’s a factor of 10^12 difference.

• Max voltage is 100 volts, and max field is 10000 N/C, which is the same as 10000 V/m

100 V / 10000 V/m = 0.01 meter, max spacing (2)

Parallel plate cap

ε₀ is 8.8542e-12 F/m

εᵣ is dielectric constant (vacuum = 1)

A and d are area of plate in m² and separation in m

5e-12 = (8.85e-12) (A) / (0.01)

A = 0.00565 m² = πr²

r = 0.0424 m or 4.24 cm (1)

Q = CV = 5pF x 100v = 500 pC (3)

I think you are screwing up the units and exponents. The last part is just a simple multiplication, and you got it wrong.

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