A 8.31-pF, parallel-plate, air-filled capacitor with circular plates is to be us

Question

A 8.31-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 2.00 times 10^2 V. The electric field between the plates is to be no greater than 1.00 times 10^5 N/C. As a budding electrical engineer for Live-Wire Electronics, your tasks are the following. (a) Design the capacitor by finding what its physical dimensions and separation must be. radius cm plate separation cm (b) Find the maximum charge these plates can hold. pC

Explanation / Answer

a)

here

d = V / E

d = 200 / (100000) = 0.002 m

the seperation between plates is 0.002 m

then

C = e0 * A / d

A = C * d / e0 = (8.31 * 10^-12 * 0.002 ) / ( 8.85 * 10^-12 )

A = 1.87 * 10^-3 m^2

then

A = pie * r^2 = 1.87 * 10^-3

3.14 * r^2 = 1.87 * 10^-3

r = sqrt( (1.87 * 10^-3 ) / 3.14)

r = 0.024 m

b)

Q = C * V

Q = 8.31 * 10^-12 * 200 = 831 pC

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.