Two identical parallel plate air capacitors are connected in series to a battery of emf& V . If one of the capacitor is completely filled with dielectric material of constant K , then potential difference of the other

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Correct answer: A. (KV)/( (K + 1))

Both the capacitance will have same charge

{V}_{1} = \dfrac{q}{C}

and

{V}_{2} = \dfrac{q}{KC}

But

{V}_{1} + {V}_{2} = V

⇒ \dfrac{q}{C} + \dfrac{q}{KC} = V

⇒ \dfrac{q}{C} . \dfrac{K + 1}{K} = V

Potential difference across capacitor without dielectric is

V ' = \dfrac{q}{C} = \dfrac{K}{K + 1} V

Two identical parallel plate air capacitors are connected in series...

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