Figure shows two identical capacitors, C1 and C2, each of 1 ?F capacitance connected to a battery of 6 V. Initially switch 'S' is closed. After sometime 'S' is left open and dielectric slabs of dielectric constant K = 3 are inserted to fill completely the space between the plates of the two capacitors. How will the (i) charge and (ii) potential difference between the plates of the capacitors be affected after the slabs are inserted ? Capacitors are connected in parallal
Asked by Anjali Raj | 21st Jun, 2013, 02:33: PM
Since, the capacitors are connected in parallel, the potential difference across each capacitor would be same and equal to that offered by the battery = 6V.
Initially, in the absence of any dielectric, C1 = C2= 1 ?F
Hence, Q1 = C1*V = 6 C
Q2 = C2*V = 6 C
After the dielectric is inserted, then C1' = C2' = KC1 = KC2 = 3*1 = 3?F
The potential difference would still remain the same, that provided by the battery.
Hence, Q1' = C1'*V = 3*6 = 18C
Q2' = C2' *V = 3*6 = 18C
Hence, after the dielectric slabs are inserted between the capacitors, the charge on each capacitor would increase by 6C, however the potential difference would still remain the same as that would be equivalent to that provided by the battery.
Answered by | 23rd Jun, 2013, 09:05: AM
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