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Two capacitors of capacitance 10muF" and...

Two capacitors of capacitance `10muF" and "20muF` are connected in series with a 6V battery. After the capacitors are fully charged, a slab of dielectric constant k is inserted between the plates of the two capacitors. How will the following be affected after the slab is introduced?
The charges on the two capacitors.

Text Solution

Verified by Experts

Here, equivalent capacitance,
`C_(eq)= (C_1C_2)/(C_1+C_2)= (10xx10^(-6)xx20xx10^(-6))/((10+20)xx10^(-6))`
`= 6.67xx10^(-6)F`
Hence, the charge becomes `Q= C_(eq)V = 6.67xx10^(-6)xx6= 4xx10^(-5)C`
Now, `V_1 = (Q)/(C_1)= (4xx10^(-5))/(10xx10^(-6)) = 4V`
`V_2 = (Q)/(C_2) = (4xx10^(-5))/(10xx10^(-6))= 2V`
After the dielectric slab is introduced, the capacitance becomes C' = kC
Now, Q= CV
`:. Q' = C'V= kCV = kQ`
Hence, `Q_1'= Qk C =(4xx10^(-5))kC`
`Q_2' = (4xx10^(-5))kC`
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