Question
A parallel plate capacitor with air as dielectric is charged by a d.c. source to a potential ‘V’. Without disconnecting the capacitor from the source, air is replaced by another dielectric medium of dielectric constant 10. State with reason, how does (i) electric field between the plates and (ii) energy stored in the capacitor change.
Solution
i) When, capacitor with air as dielectric is replaced by another dielectric medium of dielctric constant 10, without disconnecting the capacitor from the d.c source, potential difference between the capacitor plates remain constant.
Electric field between the plates is given by .
Since, there is no change in the potential difference and seperation of plates hence, electric field between the plates will remain constant.
ii) After the dielectric medium is placed in between the plates, capacitance will increase by a factor of 10.
Before, electrostatic energy is given by U0=
After, electrostatic energy is given by
Therefore, energy stored in the capacitor becomes 10 times of the initial energy.
Electric field between the plates is given by .
Since, there is no change in the potential difference and seperation of plates hence, electric field between the plates will remain constant.
ii) After the dielectric medium is placed in between the plates, capacitance will increase by a factor of 10.
Before, electrostatic energy is given by U0=
After, electrostatic energy is given by
Therefore, energy stored in the capacitor becomes 10 times of the initial energy.
