Question

Two parallel plate capacitors with capacitances $C$ and $2C$ are joined in parallel and the combination is connected to a battery of potential difference $V$. If the battery is now removed and a material of dielectric constant $K$ is filled between the plates of capacitor $C$ completely, then what will be the potential difference across the two capacitors?

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

Answer 1

The correct option is C $\frac{3V}{K+2}$


Let the initial potential difference be $V$

So, the initial charge on first capacitor, $q_1=CV$

Initial charge on second capacitor, $q_2=2CV$

As the battery is disconnected and a dielectric is introduced, the charge will redistribute on the capacitors.

However, as this is an isolated system, charge will remain conserved.

Let, the new potential difference be $V^{\prime }$ which will be the same for both capacitors as they are connected in parallel.

If $q_1^{\prime }$ and $q_2^{\prime }$ are the final charges on the capacitors, then from the conservation of charge

$q_1+q_2=q_1^{\prime }+q_2^{\prime }$

$\Rightarrow CV+2CV=KC{V}^{\prime }+2C{V}^{\prime }$

$\Rightarrow V^{\prime }=\frac{3V}{K+2}$

Hence, option (c) is the correct answer.