Question

A capacitor of capacitance C=900 pF is charged rally by 100 V battery B as shown in figure (a). Then B is disconnected from the battery and connected to another uncharged capacitor of capacitance C=900 pF as shown in figure (b). The electrostatic energy stored by the system (b)is:

• A. \(4.5 \times 10^{-6} J\)
• B. \(3.25 \times 10^{-6} J\)
• C. \(2.25 \times 10^{-6} J\)
• D. \(1.5 \times 10^{-6} J\)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
When a charged capacitor is connected to an uncharged one, the total charge is conserved but distributed across both.
Due to this redistribution, some energy is lost (mainly as heat and radiation), and the final energy is less than the initial energy.
Key Formula or Approach:
1. Initial Energy: \(U_i = \frac{1}{2}CV^2\)
2. Common Potential: \(V_c = \frac{C_1V_1 + C_2V_2}{C_1 + C_2}\)
3. Final Energy: \(U_f = \frac{1}{2}(C_1 + C_2)V_c^2\)
Step 2: Detailed Explanation:
1. Initial Energy calculation:
\(C = 900 \text{ pF} = 900 \times 10^{-12} \text{ F}\), \(V = 100 \text{ V}\).
\[ U_i = \frac{1}{2} \times 900 \times 10^{-12} \times (100)^2 = 450 \times 10^{-12} \times 10^4 = 4.5 \times 10^{-6} \text{ J} \]
2. Common Potential calculation:
Since \(C_1 = C_2 = 900 \text{ pF}\) and the second is uncharged:
\[ V_c = \frac{C \times 100 + C \times 0}{C + C} = \frac{100C}{2C} = 50 \text{ V} \]
3. Final Energy of system (b):
The total capacitance is \(C_{total} = C + C = 1800 \text{ pF}\).
\[ U_f = \frac{1}{2} \times (1800 \times 10^{-12}) \times (50)^2 \]
\[ U_f = 900 \times 10^{-12} \times 2500 = 225 \times 10^4 \times 10^{-12} = 2.25 \times 10^{-6} \text{ J} \]
Step 3: Final Answer:
The energy stored by the system is \(2.25 \times 10^{-6}\) J.