Two capacitors C1=8μF and C2=4μF are connected in series between points A and B. An ideal cell of emf 12V is connected between A and B for a long time. Now, a slab of dielectric constant k=2 is slowly introduced, fully in the gap of capacitor C2.
A
As the dielectric is introduced, the energy taken from the cell is 192μJ.
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B
As the dielectric is introduced the increase in electrostatic energy stored in the capacitors is 96μJ.
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C
The work done by dielectric slab on the filling agent during the filling process is positive.
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D
The work done by dielectric slab on the filling agent during the filling process is zero.
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Solution
The correct options are A As the dielectric is introduced, the energy taken from the cell is 192μJ. B As the dielectric is introduced the increase in electrostatic energy stored in the capacitors is 96μJ. C The work done by dielectric slab on the filling agent during the filling process is positive. Before introducing the dielectric, the equivalent capacitance of the circuit is Ci=C1C2C1+C2=3212=83μF. After introducing the dielectric, the equivalent capacitance of the circuit is Cf=kC1C2C1+kC2=6416=4μF. Change in charge flowing through battery Δq=ΔCV=(4−83)12=16μC Energy taken from the cell = charge flown through the cell x cell voltage =16×12=192μJ.
Electrostatic energy stored in capacitors initially is Ui=12CV2=(12)×83×122=192μJ. Electrostatic energy stored in capacitors after introducing dielectrc is Uf=12CV2=(12)×4×122=288μJ. Therefore, increase in the electrostatic energy stored in the capacitors =96μJ.