A 100pF capacitor is charged to a potential difference of 24V. it is c...
Explanation:
Step 1: Initial Charge and Potential Difference
- The initial charge on the 100pF capacitor is Q = C1V1 = (100x10^-12) x 24 = 2.4 x 10^-9 C
- The potential difference across the 100pF capacitor is V1 = 24V
Step 2: Charge Sharing
- When the 100pF capacitor is connected to the uncharged 20pF capacitor, the total capacitance becomes C = C1 + C2 = 120pF
- The charge on the 100pF capacitor will now be shared between the two capacitors, with some charge flowing from the 100pF capacitor to the 20pF capacitor until they both have the same potential difference.
Step 3: Final Charge and Potential Difference
- Let Q1 be the final charge on the 100pF capacitor and Q2 be the final charge on the 20pF capacitor.
- The total charge remains the same, so Q1 + Q2 = 2.4 x 10^-9 C.
- The potential difference across each capacitor is now the same, say V.
- Using the formula Q = CV, we can write:
Q1 = C1V, Q2 = C2V
Q1 + Q2 = (C1 + C2)V
V = Q / C = (Q1 + Q2) / (C1 + C2)
V = (2.4 x 10^-9) / (120 x 10^-12) = 20V
Step 4: Conclusion
- Therefore, the final potential difference across both capacitors is V = 20V.
- The energy stored in the system is proportional to the square of the potential difference, so the energy stored in the capacitors has decreased by a factor of (20/24)^2 = 0.69.