MHT CET · Physics · Capacitance

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 capacitor will become

A $\frac{K V}{(K + 1)}$
B $\frac{V}{K + 1}$
C $\frac{(K - 1) V}{K}$
D $\frac{V}{K (K + 1)}$

Verified Solution

Answer & Solution

Correct Answer

(A) $\frac{K V}{(K + 1)}$

Step-by-step Solution

Detailed explanation

Both the capacitance will have same charge

V_{1} = \frac{q}{C}

and

V_{2} = \frac{q}{K C}

But

V_{1} + V_{2} = V

\Rightarrow \frac{q}{C} + \frac{q}{K C} = V

\Rightarrow \frac{q}{C} . \frac{K + 1}{K} = V

Potential difference across capacitor without dielectric is

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

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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