Initially, the two capacitors are charged up to the same voltage V, so the charge on the capacitor of capacitance C is Q = CV, and the charge on the capacitor of capacitance 2C is Q' = 2CV.
When the battery is removed and a dielectric medium of constant K is inserted between the plates of the first capacitor (of capacitance C), the capacitance of this capacitor becomes KC. The capacitance of the second capacitor remains the same at 2C.
The total charge on the capacitors remains the same, as no charge is lost or gained during this process. Therefore, we have:
Q = Q' = 2CV
After the insertion of the dielectric medium, the potential difference across the two capacitors will be different, as their capacitances have changed. Let V1 be the new potential difference across the capacitor of capacitance KC, and V2 be the potential difference across the capacitor of capacitance 2C.
We know that the total charge on the capacitors remains the same, so we can write:
Q = C1V1 + C2V2
where C1 = KC is the capacitance of the first capacitor (after insertion of the dielectric), and C2 = 2C is the capacitance of the second capacitor.
Substituting the values of C1, C2 and Q, we get:
CV = KV1 + 2CV2
Simplifying, we get:
V1 = (C/K) V - 2V2
But we also know that the potential difference across the two capacitors must be the same, i.e., V = V1 + V2. Substituting the value of V1 in this equation, we get:
V = [(C/K) V - 2V2] + V2
Simplifying, we get:
V2 = (1/3) V/K
Therefore, the potential difference across the capacitor of capacitance KC (after insertion of the dielectric) is:
V1 = (C/K) V - 2V2 = (5/3) V - (2/3) V/K
Thus, the potential on each capacitor after the insertion of the dielectric is V1 = (5/3) V - (2/3) V/K for the capacitor of capacitance KC, and V2 = (1/3) V/K for the capacitor of capacitance 2C.