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Step-by-Step Detailed Solution
Step 1: Identify the given information
• There are two capacitors, one of capacity $2C$ and the other of capacity $C$.
• They are first connected in parallel and charged to a potential difference $V$.
• After charging, the battery is removed.
• The capacitor of capacity $C$ is then completely filled with a dielectric medium of dielectric constant $K$.
Step 2: Compute the total initial charge
Since the two capacitors are in parallel and charged to $V$, their combined charge is the sum of individual charges:
$Q_{\text{total}} \;=\;(2C)\,V \;+\; (C)\,V \;=\; 3CV$.
Step 3: Express the new effective capacitances
• The capacitor previously of capacity $2C$ remains unchanged (still $2C$).
• The capacitor of capacity $C$ is now filled with a dielectric of constant $K$, so its new capacitance becomes $KC$.
Step 4: Apply charge conservation to find the new potential
After inserting the dielectric and ensuring the capacitors remain connected in parallel (with the battery removed), the total charge $Q_{\text{total}} = 3CV$ redistributes across the new combination. Since both capacitors are in parallel, the final potential $V_f$ is the same across each capacitor. Thus:
$ Q_{\text{total}} = (2C)\,V_f \;+\; (KC)\,V_f \;=\; (2C + KC)\,V_f.$
Therefore,
$ V_f \;=\; \dfrac{Q_{\text{total}}}{2C + KC} \;=\; \dfrac{3CV}{2C + KC} \;=\; \dfrac{3V}{K + 2}.$
Step 5: Conclude the final potential difference
Hence, the potential difference across the capacitors after the dielectric is introduced is:
$V_f = \dfrac{3V}{K + 2}$.
Final Answer: $ \dfrac{3V}{K + 2} $
Reference Image and Expression
$V_C = \dfrac{2CV + CV}{KC + 2C} = \dfrac{3V}{K + 2}$
