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

Question

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, 3 . 25 × 10 - 6 J, 2 . 25 × 10 - 6 J, 1 . 5 × 10 - 6 J, 4 . 5 × 10 - 6 J

MARKS App · Accepted Answer

Initial charge on 900 pF capacitor will be q = C V When the two equal capacitors are joined together as shown in figure (b). The charge will be equally divided among them. Therefore, q ' = q 2 = C V 2 Now the energy stored in them will be E = q ' 2 2 C + q ' 2 2 C ⇒ E = C V 2 2 × 1 C = C V 2 4 = 900 × 10 - 12 × 100 2 4 ⇒ E = 2 . 25 × 10 - 6 J

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