Two identical parallel plate air capacitors are connected in series to a battery of emf $V$. If one of the capacitors is inserted in a liquid of dielectric constant $K$,then the potential difference of the other capacitor will become:
- A$\frac{K-1}{KV}$
- B$\frac{K+1}{KV}$
- C$\frac{KV}{K+1}$
- D$\frac{KV}{K-1}$
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