Question

8. A parallel plate capacitor, each with platea
and separation d, is charged to a potential
difference V. The battery used to charge it rermain
connected. A dielectric slab of thickness d
dielectric constant k is now placed between
plates. What change, if any, will take place
charge on plates? (it) electric field intensity between the plates ?
capacitance of capacitor?
justify your answer in each case​

Answer 1

C=Aε/d

Q=CV

=Aε/d*V

new C=kC

C'=kAε/d

Q'=kCV

electric field intensity=σ/ε

=Q'/A/ε

=Q'/Aε

=kCV/Aε