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
Answers
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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
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