Question

Two parallel plate capacitors of capcitances C and 2C are connected in parallel and changed to a potential V by a battery. The battery is then disconnected and the apsce between the plates of capacitor C is completely filled with a material of dielectric constant K. The potential difference across the capacitors now becomes

• A. $\frac{V}{K+1}$
  
• B. $\frac{2V}{K+2}$
  
• C. $\frac{3V}{K+2}$
  
• D. $\frac{3V}{K+3}$

Answer 1

The correct option is C

$\frac{3V}{K+2}$

Original capacitance of the parallel combination of C and 2C =C+2C=3C. So, total charge Q=3CV.
When the capacitor C is filled with dielectric, its capacitance becomes KC. Therefore, the capacitance of the combination after the capacitor C is filled with dielectric, C'=KC+2C=(K+2)C.
Since, the charge remains the same, Q=3CV, the potential difference across the capacitors will be
$\frac{Q}{C^{\prime }}=\frac{3CV}{(K+2)C}=\frac{3V}{K+2}$