1.Find the capacitance of a parallel plate capacitor having plates of area 5.00 m2 that are separated by 0.200 mm of Teflon.
2.What capacitance is needed to store 3.00 µC of charge at a voltage of 120 V?
Here ,
area of plates , A = 5 m^2
seperation of plates , d = 0.2 mm
dielectric constant of teflon , k = 2.1
for a parallel plate capacitor is given as
C = k*epsilon*A/d
C = 2.1 * 8.854 *10^-12 * 5/(.2 *10^-3)
C = 4.65 *10^-7 F
the capacitanc is 4.65 *10^-7 F
2)
as charge = capaitance * V
3 *10^-6 = C * 120
C = 2.5 *10^-8 F
the capacitance is 2.5 *10^-8 F
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