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Figure shows two identical capacitors C1...

Figure shows two identical capacitors `C_1 and C_2` each of `2muF` capacitance, connected to a battery of 5 V. Initially switch 'S' is closed. After some time 'S' is left open and dielectric slavs of dielectric constant K = 5 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) When S is open charge on each capacitor will be
`{:(q_1=C_1V,q_2=C_2'V),(q_1=5CV=5xx2xx5=50muC,q_2=50muC0^(-6)xx2):}`
(ii) P.d. across `C_1` is still 5V and across `C_2, q=(5C)V`
`V'=V/5=5/5=1V`.
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