The parallel combination of two air filled parallel plate capacitors of capacitance $C$ and $nC$ is connected to a battery of voltage, $V.$ When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant $K$ is placed the two plates of the first capacitor. The new potential difference of the combined system is:

• Correct answer
  A

  $\frac{\left(n+1\right)V}{\left(K+n\right)}$
• B

  $\frac{nV}{K+n}$
• C

  $\frac{V}{K+n}$
• D

  $V$

Solution

i)

ii) Battery disconnected after full charge,
Total charge of the system,
$=CV+nCV$
$=\left(n+1\right)CV$

iii) Inserting dielectric of constant $K$

Common potential, $V_C=\frac{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}}{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{c}\mathrm{a}\mathrm{p}\mathrm{a}\mathrm{c}\mathrm{i}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}}$
$=\frac{\left(n+1\right)CV}{KC+nC}=\frac{\left(n+1\right)V}{K+n}$

Hence, the correct answer is option: 'A'.