A parallel-plate capacitor, of area 3.56×10 -2 m 2 , has the gap between the pla
ID: 1546945 • Letter: A
Question
A parallel-plate capacitor, of area 3.56×10-2 m2, has the gap
between the plates filled by Plexiglas, a dielectric. A laboratory meter measures the capacitance as 510 pF.
What is the capacitance when there is only air in the gap?
What is the separation of the plates?
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Answer 1 of 2:
Answer 2 of 2: Last Answer 1: 2.43*10^2 pF Last Answer 2: 1.3 mm
Incorrect, tries 3/15.
A parallel-plate air capacitor is made from two plates 0.320 m2 in area, spaced 3.30 cm apart. The potential difference between the plates is 59.0 V.
What is the charge on each plate?
What is the electric field between the plates?
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Answer 1 of 2:
Answer 2 of 2:
Explanation / Answer
given:
1. area =3.56×10-2 m2
capacitance = 510 pF = 510*10^-12 F
plexiglass dielectric constant = 3.40
air dielectric constant =1
capacitance C1 = k1*(epsilon 0) A/d------------(1)when there is dielectric
capacitance C2 = k2*(epsilon 0) A/d------------(2)when there is air
(1)/(2)
C1 / C2 = k1/k2
(510*10^-12)/(C2) =3.40/1
C2 = 1.5*10^-10 F =150 pF
separation between the plates d=?
C1 = k1*(epsilon 0) A/d
510*10^-12 = (3.40*8.85*10^-12*3.56*10^-2)/d
d = 0.0021m = 0.21cm
2.are = 0.320m^2
d = 3.30cm = 0.0330m
V = 59V
capacitance C = k*(epsilon 0) A/d
C = (1*8.85*10^-12*0.320)/(0.0330) = 8.58*10^-11 F = 85.8pF
E = V/d = 59/(8.58*10^-11) = 0.687*10^12 V/m
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