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A capacitor of capacitance C = 900 pF is charged fully by a 100 V battery B as shown in Figure (a). Then it 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

1.5 × 10⁻⁶ J

B

4.5 × 10⁻⁶ J

C

3.25 × 10⁻⁶ J

D

2.25 × 10⁻⁶ J

Step-by-Step Solution

  1. Initial State (Figure a):
  • Capacitance C1=900 pF=9×10−10 FC_1 = 900 \text{ pF} = 9 \times 10^{-10} \text{ F}.
  • Voltage V1=100 VV_1 = 100 \text{ V}.
  • Initial Energy stored: Ui=12C1V12=12(9×10−10)(100)2=4.5×10−6 JU_i = \frac{1}{2}C_1 V_1^2 = \frac{1}{2}(9 \times 10^{-10})(100)^2 = 4.5 \times 10^{-6} \text{ J}.
  • Charge stored: Q=C1V1=(900×10−12)(100)=9×10−8 CQ = C_1 V_1 = (900 \times 10^{-12})(100) = 9 \times 10^{-8} \text{ C}.
  1. Final State (Figure b):
  • The charged capacitor C1C_1 is disconnected and connected in parallel to an identical uncharged capacitor C2=900 pFC_2 = 900 \text{ pF}.
  • Charge Redistribution: The total charge QQ remains conserved but distributes equally between the two identical capacitors. Thus, charge on each is Q′=Q/2Q' = Q/2.
  • System Capacitance: The two capacitors are in parallel, so equivalent capacitance Ceq=C1+C2=2C=1800 pFC_{eq} = C_1 + C_2 = 2C = 1800 \text{ pF}.
  • Final Energy: The energy stored in the system is Uf=Q22CeqU_f = \frac{Q^2}{2C_{eq}}.
  • Substituting Ceq=2C1C_{eq} = 2C_1: Uf=Q22(2C1)=12(Q22C1)=12UiU_f = \frac{Q^2}{2(2C_1)} = \frac{1}{2} \left( \frac{Q^2}{2C_1} \right) = \frac{1}{2} U_i.
  1. Calculation:
  • Uf=12×(4.5×10−6 J)=2.25×10−6 JU_f = \frac{1}{2} \times (4.5 \times 10^{-6} \text{ J}) = 2.25 \times 10^{-6} \text{ J}.
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