Question

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 1.00×102v. The electric field between the plates is to be no greater than 1.00×104N/C. As a budding electrical engineer for Live Wire Electronics, your tasks are to (a) design the capacitor by finding what its physical dimensions and separation must be; (b) find the maximum charge these plates can hold.

Answer 1

a) r=4.24cm d=1 cm

b)
`Q=5x10^(-10) C`

Step-by-step explanation:

The capacitance depends only of the geometry of the capacitor so to design in this case knowing the Voltage and the electric field

`V=1.00x10^(2)v\\E=1.00x10^(4) (N)/(C)`

`V=E*d\\d=(V)/(E)\\d=(1.0x10^(2))/(1.0x10^(4))\\d=0.01m`

The distance must be the separation the r distance can be find also using

`C=(Q)/(V_(ab))`

But now don't know the charge these plates can hold yet so

a).

d=0.01m

`C=E_(o)*(A)/(d)\\A=(C*d)/(E_(o))`

`A=(5pF*0.01m)/(8.85x10^(-12)(F)/(m))\\A=5.69x10^(-3)m^(2)`

`A=\pi *r^(2)\\r=\sqrt{(A)/(r)}\\r=\sqrt{(5.64x10^(-3)m^(2) )/(\pi ) }  \\r=42.55x^(-3)m`

b).

`C=(Q)/(V_(ab))`

`Q=C*V\\Q=5x10^(-12) F*1x10^(2)\\Q=5x10^(-10)C`