(14) A parallel combination of two parallel-plate air-gap capacitors, each havin
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Question
(14) A parallel combination of two parallel-plate air-gap capacitors, each having a capacitance of 2 F are connected in parallel across a 12.0-V battery. The battery is disconnected from the parallel combination, and then a slab that has a dielectric constant 2.5 is inserted between the plates of one of the capacitors, completely filling the gap. Before the dielectric slab is inserted, find (a) the charge on and energy stored in each capacitor, and (b) the total energy stored in the capacitors. After the dielectric is inserted, find c) the potential difference across each capacitor, (d) the charge on each capacitor, (e) the total energy stored in the capacitorsExplanation / Answer
a) since they are in parallel potential across both will be same as the source
Q= CV= 12*2*10-6
Q=24*10-6 Coloumbs on each capacitor
energy stored is, U= 0.5*C*V2 = 0.5*2*10-6*12*12
U= 1.44*10-4 J on each capacitor
b) total energy is 2.88*10-4 J
c) potential difference across unchanged capacitance will be same as 12V
Across new capacitance will be, V= Q/C
Since capacitance will be increased by a factor of k. Hence potential will be decreased by a factor of k
Potential across new capacitance is ,V = 12/2.5= 4.8V
d) charge will remain same on each capacitor.
As in part a
e)energy on one capacitor is 1.44*10-4 J
On second capacitor is 0.5*2.5*2*10-6*4.8*4.8= 5.76*10-5 J
Total energy is 2.106*10-4 J
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