A parallel-plate capacitor consists of plates of area 1.5 x 10 -2 m 2 that are s

Question

A parallel-plate capacitor consists of plates of area 1.5 x 10-2 m2 that are separated by 1.0 mm. The capacitor is connected to a 12 V battery.

a) what is the charge on the plates?

b) How much energy is stored in the plates?

c) once the plates are fully charged, the battery is removed and a slab of dielectric that completely fills the space between the plates is inserted. if the dielectric has a dielectric constant of 5.0, what is the capacitance of the capacitor after the slab is inserted?

d) what is the voltage across the capacitor's plates after the slab is inserted?

e) this capacitor (with the dielectric) is now connected to a second capacitor of capacitance 15 pF. What is the charge across each capacitor after they are connected?

Explanation / Answer

Area A = 1.5 x 10-2 m2
Separation d = 1.0 mm                     = 10 -3 m Voltage V = 12 V a) The charge on the plates Q = CV Where C = capacitance                 = A/d              = permitivity of free space                 = 8.85 x 10 -12 C 2 / N m 2 Substitue values we get C = 1.3275 x 10 -10 F Q = 1.593 x 10 -9 coulomb's         = 1.593 nC b) Energy is stored in the plates E = ( 1/ 2) CV 2                                                    = 9.558 x 10 -9 J c) Dielectric constant k = 5.0 The capacitance of the capacitor after the slab is inserted C ' = k C                                                                                             = 6.6375 x 10 -10 F d) The voltage across the capacitor's plates after the slab is inserted V ' = V / k                                                                                                             = 2.4 volt

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