# A parallel-plate air-filled capacitor has a capacitance of 43 pF.?

A parallel-plate air-filled capacitor has a capacitance of 43 pF. (a) If each of its plates has an area of 0.28 m2, what is the separation? (b) If the region between the plates is now filled with material having κ = 8.2, what is the capacitance?

### 3 Answers

- JacobLv 59 years agoFavorite Answer
43 pF = 0.000043 microF

using the equation C=(k*epilison*Area)/separation

where,

epsilon = 8.854*10^-12 F/m

C= 43 pF

A=0.28

k=1.0

solve for d (separation)

d=5.765 cm

b)

C = (8.2*0.28*8.854*10^-12)/(0.05765)

C = 3.525*10^-10 F = 352.5 pF

- 4 years ago
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A parallel-plate air-filled capacitor has a capacitance of 43 pF.?

A parallel-plate air-filled capacitor has a capacitance of 43 pF. (a) If each of its plates has an area of 0.28 m2, what is the separation? (b) If the region between the plates is now filled with material having κ = 8.2, what is the capacitance?

Source(s): parallel plate air filled capacitor capacitance 43 pf: https://shortly.im/7KE7V - CynthiaLv 44 years ago
C = 1.6 pF = (epsilon*Area*k)/d rearranging, d = (epsilon*Area*k)/1.6pF where, k=1.0 epsilon = 8.854*10^-12 F/m area = unknown doubling the separation, d = (2*area*1.0*8.854*10^-12)/(1.6*10^-12) d = 11.0675Area Now that d will be plugged into the new equation 2.8 pF = (k*area*epsilon)/d 2.8 pF = (k*area*epsilon)/(11.0675Area) areas cancel out, and you solve for k k = 3.5