A first capacitor is initially connected to potential source ?. The charged capacitor is then removed
from the source and connected to a second initially uncharged capacitor. Determine the final potential
across the first capacitor long after the switch is closed.
2;1
To determine the final potential across the first capacitor, we can use the concept of charge conservation.
1. Let's assume that the initial potential across the first capacitor (C1) is V1 and it carries a charge of Q1.
2. Since the capacitor is initially connected to a potential source, it will get charged up to its maximum potential which is equal to the potential source.
3. Now, when the charged capacitor (C1) is removed from the source and connected to the second uncharged capacitor (C2), the charge on C1 will start redistributing between C1 and C2.
4. Let's assume that the charge on C1 redistributes to Q1' and the charge on C2 becomes Q2.
5. The total charge remains conserved. Therefore, Q1 + Q2 = Q1' + Q2.
6. The potential difference across each capacitor is given by the formula V = Q/C, where V is the potential difference, Q is the charge, and C is the capacitance.
7. Since the capacitance of C1 remains constant, the initial potential across C1 (V1) is equal to the initial potential difference (V1 = Q1/C1).
8. Similarly, the potential difference across C2 is V2 = Q2/C2, where Q2 is the charge on C2 and C2 is the capacitance of C2.
9. The final potential across C1 is the potential difference when the redistribution of charge is complete, which means Q1' = Q1.
10. Thus, the final potential across C1 is V1' = Q1'/C1 = Q1/C1 = V1 (since Q1 = Q1').
11. Therefore, the final potential across the first capacitor (C1) long after the switch is closed remains the same as the initial potential, which is V1.
So, the final potential across the first capacitor remains unchanged.