Three capacitor A,B and C are connected with battery of emf $ε$ . All capacitors are identical initially.

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If capacitor C is removed from the circuit, (assuming without any work) then Chemical energy of battery remains unchanged

If separation between plated of capacitor A is slowly increased by external force, then Force between plates of capacitor C decreases.

If dielectric slab is inserted between plates of capacitor B slowly, force between plates of capacitor C decreases.

If dielectric slab is inserted between plates of capacitor A slowly with help of external force, magnitude of change of charge on B and C is same.

answer is A, B, D.

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Detailed Solution

Initial charge distribution is shown in figure
$Q = \frac{2 C}{3} ε$
Question Image
$C_{e q u i v a l e n t}$ Increases, charge increases and $q_{1} & q_{2}$ change of charge on B and C, then
$\frac{\frac{Q}{2} + q_{1}}{k C} = \frac{\frac{Q}{2} + q_{2}}{C} \Rightarrow \frac{Q}{2} + q_{1} = \frac{k Q}{2} + k q_{2} q_{1} \neq q_{2}$
and according to loop law
$\frac{Q + q_{1} + q_{2}}{C} + \frac{\frac{Q}{2} + q_{2}}{C} = ε$
on solving $q_{2} = - v e$
$q = + v e$
Capacitance of A increases charge increases
No change in charge.
If separation between plates of capacitor A is increased then capacitance of A is decreased $C_{e q u i v a l e n t}$ decreases, charge decreases.