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

A capacitor of capacitance C=900 pF is charged fully by 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

  1. 3.25×10-6 J
  2. 2.25×10-6 J
  3. 1.5×10-6 J
  4. 4.5×10-6 J

Solution

Initial charge on 900 pF capacitor will be q=CV

When the two equal capacitors are joined together as shown in figure (b). The charge will be equally divided among them. Therefore,

q'=q2=CV2

Now the energy stored in them will be

E=q'22C+q'22C

⇒E=CV22×1C=CV24=900×10-12×10024

⇒E=2.25×10-6 J

Asked in: NEET 2022 (Phase 1)

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