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In the adjoining figure, capacitor (1) a...

In the adjoining figure, capacitor `(1)` and `(2)` have a capacitance 'C' each. When the dielectric of dielectyric constant `K` is inserted between the plates of one of the capacitor, the total charge flowing through battery is

A

`(KCE)/(K+1)` form `B` to `C`

B

`(KCE)/(K+1)`from `C` to `B`

C

`((K-1)CE)/(2(K+1))` form `B` to `C`

D

`((K-1)CE)/(2(K+1))`from `C` to `B`

Text Solution

Verified by Experts

The correct Answer is:
D
`Q = (C )/(2) xxE`
Now `C_(eq) = (KC)/((K +1))`

`Q = ((KC)/(K+1)).E = (KCE)/(K +1)`
Charge flow thorugh battery

`=(KCE)/(K+1) -(CE)/(2) = ((K-1))/(2(K+1))CE` form `C` to `B`
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