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You need to design a capacitor capable of storing 4.0 10-7 C of charge. At your disposal, you have a 100 V power supply and two metal plates, each of area 0.240 m2. What is the limit of the separation of the plates?
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Added Fri, 03 Jul '15
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Capacitance = Q/V

(4e-7)/100 = 4e-9

C = (EoA)/d --> (EoA)/C = d

(8.85e-12).24/4e-9

d=.531mm

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Added Thu, 02 Mar '17
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Capacitance C of a parallel plate capacitor is given by C = KeoA/d
where A = area = pi r^2,
e0 = constant = 8.85*10^-12,
d   = distance between the plates,
K = dielectric constant (=1 for air)

so d = e0A/C
d = 8.85*10^-12 * 0.240/4*10^-7
d = 5.31 um
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Added Fri, 03 Jul '15
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