Answer:
The electrostatic energy stored in the system after connecting the two capacitors is 0.09 Joules
Step-by-step explanation:
The formula to calculate the electrostatic energy stored in a capacitor is given by:
E = (1/2) * C * V^2
where:
E is the electrostatic energy,
C is the capacitance, and
V is the voltage across the capacitor.
(a) For the first scenario, where a 900 pF capacitor is charged by a 100 V battery:
C = 900 pF = 900 * 10^(-12) F
V = 100 V
Using the formula, we can calculate the electrostatic energy stored in the capacitor:
E = (1/2) * C * V^2
E = (1/2) * (900 * 10^(-12)) * (100^2)
E = 0.045 J
Therefore, the electrostatic energy stored by the capacitor in the first scenario is 0.045 Joules.
(b) In the second scenario, when the first capacitor is disconnected from the battery and connected to another 900 pF capacitor, the total capacitance in the system becomes:
C_total = C1 + C2
C_total = 900 pF + 900 pF
C_total = 1800 pF = 1800 * 10^(-12) F
The voltage across the capacitors remains the same, as they are connected in parallel.
Using the formula for electrostatic energy, we can calculate the new energy stored in the system:
E_total = (1/2) * C_total * V^2
E_total = (1/2) * (1800 * 10^(-12)) * (100^2)
E_total = 0.09 J
Therefore, the electrostatic energy stored in the system after connecting the two capacitors is 0.09 Joules.