1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# 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.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
As the dielectric is introduced the increase in electrostatic energy stored in the capacitors is 96μJ.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
The work done by dielectric slab on the filling agent during the filling process is positive.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
The work done by dielectric slab on the filling agent during the filling process is zero.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is 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.

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Capacitors in Circuits
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program