Question

The energy stored in the capacitor as shown in Fig. (a) is \(4.5\times 10^{-6}\) J. If the battery is replaced by another capacitor of 900 pF as shown in Fig. (b), then the total energy of system is

• A. \(4.5\times 10^{-6}\) J
• B. \(2.25\times 10^{-6}\) J
• C. zero
• D. \(9\times 10^{-6}\) J

Hint:

When a charged capacitor is connected to an identical uncharged capacitor, total energy becomes half because some energy is lost as heat during charge redistribution.

Answer 1

The Correct Option is B

Step 1: Understand initial setup in Fig. (a).
In Fig. (a), a single capacitor \(C = 900pF\) is connected to a battery of \(100V\).
Energy stored is given:
\[ U_0 = 4.5\times 10^{-6} \, J \]
Step 2: Charge stored initially.
\[ Q = CV \]
So charge on capacitor:
\[ Q = C(100V) \]
Step 3: In Fig. (b), battery is replaced by identical capacitor.
Now two identical capacitors (each \(900pF\)) are connected together.
Charge redistributes and final voltage becomes half, because total capacitance doubles.
Step 4: Energy after redistribution.
For equal capacitors, after connection:
Final energy becomes half of initial energy:
\[ U = \frac{U_0}{2} \]
Step 5: Calculate final energy.
\[ U = \frac{4.5\times 10^{-6}}{2} = 2.25\times 10^{-6} \, J \]
Final Answer:
\[ \boxed{2.25\times 10^{-6}\, J} \]