Question

Two capacitors, one of 5 μF and the other of 8 μF, connected in parallel, are charged with a 12 volt battery. The battery is then disconnected, and while the capacitors remain connected in parallel, a dielectric with constant 2 is inserted between the plates of the 8 μF capacitor. What is the voltage across this capacitor after the dielectric is inserted?

Answer #1

**Given**

** capacitors c1 = 5 μF and c2 = 8
μF,**

**battery potential difference v = 12 V
battery disconnected and**

**dielectric constant k = 2, inserted between the plates
of capacitor c2**

**we know that if the dielectric material of constant k
placed between the plates of capacitor , the capacitance increases
by k times**

**and potential will decreases by k times**

** and Q = C*V**

** V = Q/C**

**as the battery disconnected then the charge on each
plate is constant (does not change)**

** Q1 = c1*v = 5 μF*12 = 60 μF
Q2 = c2*v = 8 μF*12 = 96 μF**

**after the battery disconnected**

** v1 = Q1/c1 = 60/5 = 12 V**

** v2 = Q2 /(kc2) = 96/(2*8) = 6
V**

**it would be redistributed (because they are in parallel)
then the potential across each capacitor is 9 V**

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