A parallel plates capacitor is made by placing two plates of area 8.70cm 2 a dis
ID: 1385542 • Letter: A
Question
A parallel plates capacitor is made by placing two plates of area 8.70cm2 a distance of 1.65mm apart.
a) Calculate the capacitance in pF assuming that there is air in between the plates.
b) What would be the charge in pC, on each of the plates when we apply a voltage of 23.0-V to the capacitor.
c) How much energy, in Joules, is stored then in the capacitor?
d) What if we fill the space between the plates with paper whose dielectric constant ? is 9.0, what would be the new capacitance in pF?
Use 8.85 x 10-12 F/m for ?o, the permittivity of free space or air.
Explanation / Answer
a.
The capacitance of the parallel plate capacitor is,
C = ke0A/d = 1.00059*8.85x10-12*8.7x10-4m2/1.65x10-3 = 4.67x10-12 F = 4.67 pF
b.
The charge on the plate is, Q = CV = 4.67x10-12 * 23 = 107x10-12 C = 107 pC
c.
The energy stored in the capacitor is,
U = 0.5CV2 = 0.5* 4.67x10-12 * 23 *23 = 1.23x10-9 J
d.
The capacitance of the parallel plate capacitor is,
C = kC0 = 42 pF
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