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Figure shows two identical caoacitors C1...

Figure shows two identical caoacitors `C_1 and C_2` each of `1.5muF` caoacitance, connected to a battery of 2 V. Initially switch 'S' is closed. After some time 's' is left open and dielectric slabs of dielectric constant K=2 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.

Text Solution

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(i) Initially when, S is closed. `C_1 and C_2` are in parallel
`C=C_1+C_2=1.5muF+1.5muF+1.5 muF=3muF`
Total charge `Q=CV=3xx10^(-6)xx2=6xx10^(-6)C=6muC`
Charege on each capacitor is `3muC`.
V will remain same, `Q_1'=C_1'=KC_1V=2xx1.5xx10^(-6)xx2=6xx10^(-6)` and same will be `Q_2`.
(ii) Potential difference
`V=(Q_1+Q_2)/(C_1+C_2)=(6xx10^(-6))/(1.5xx10^(-6)+1.5xx10^(-6)xx2)=30/15=2V`.
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