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Two parallel plate air capacitors each of capacitance C were connected in series to a battery with emf E. Then one of the capacitors was filled up with uniform dielectric with relative permittivity k. What amount of charge flows through the battery?

Two parallel plate air capacitors each of capacitance C were connected in series to a battery with emf E. Then one of the capacitors was filled up with uniform dielectric with relative permittivity k. What amount of charge flows through the battery?

Grade:11

2 Answers

Khimraj
3007 Points
3 years ago
initial charge Qi = 2CE
final  capacitance are C and KC
equivalent capacitance = KC/(1+K)
final charge = KCE/(1+K)
charge flown through the battery = final charge – initial charge
= KCE/(1+K) – 2CE
Gautam Jain
5 Points
3 years ago
 
initial equivalent charge Qi = CE/2 
final  capacitance are C and KC
equivalent capacitance = KC/(1+K)
final charge = KCE/(1+K)
charge flown through the battery = final charge – initial charge
= KCE/(1+K) – CE/2
=1/2CE(K-1/K+1)

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