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

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$\frac{(K - 1) CE}{2 (K + 1)}$ from C to D

$\frac{KCE}{K + 1}$ from B to C

$\frac{KCE}{K + 1}$ from C to B

$\frac{(K - 1) CE}{2 (K + 1)}$ from B to C

answer is D.

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Detailed Solution

Initially,

$Q = \frac{C}{2} E$

After inserting the slab,

$Q' = \frac{KCC}{KC + C} E = \frac{KC}{K + 1} E$

$∴ Q' - Q = \frac{KCE}{K + 1} - \frac{CE}{2} = \frac{(K - 1) CE}{2 (K + 1)}$

This charge is supplied by battery.

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