Answer:
The additional energy required to be added to charge the 2μF from 50V to a 100V is 0.0075J.
Explanation:
Energy of a capacitor is given by
E = CV²/2 where C is capacitance = 2μF = 2 × 10⁻⁶ F
When the capacitor is charged to 50V, the energy it has is,
E = (2 × 10⁻⁶ × 50²)/2 = 0.0025 J
When the capacitor is charged to 100V, the energy it has stored,
E = (2 × 10⁻⁶ × 100²)/2 = 0.01 J
Energy difference between being charged to 50V and being charged to 100V is,
0.01 - 0.0025 = 0.0075 J
This represents the additional energy required to be added to charge the 2μF from 50V to a 100V.
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