Work done by 1 V battery when switch S is closed in - Jis -

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Standard XII

Physics

Capacitance of Parallel Plate Capacitor

Work done by ...

Question

Work done by 1 V battery when switch S is closed in $μ J$ is .

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Solution

Initial charge on left plate of 3 $μ$ F is -8 $μ C$ .
Charge on left plate of $3 μ$ F after closing switch is $- 3 μ C$ .
Therefore, charge flown through 1V after closing switch is $- 5 μ C$ . So, work done by the battery is $W = Q \times V = - 5 μ \times 1 = - 5 μ J$

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Work done by 1 V battery when switch S is closed in μJis .

Initial charge on left plate of 3μF is -8μC. Charge on left plate of 3μF after closing switch is −3μC. Therefore, charge flown through 1V after closing switch is −5μC. So, work done by the battery is W=Q×V=−5μ×1=−5μJ

Work done by 1 V battery when switch S is closed in $μ J$ is .

Initial charge on left plate of 3 $μ$ F is -8 $μ C$ .
Charge on left plate of $3 μ$ F after closing switch is $- 3 μ C$ .
Therefore, charge flown through 1V after closing switch is $- 5 μ C$ . So, work done by the battery is $W = Q \times V = - 5 μ \times 1 = - 5 μ J$