A charge of 2.00 μC flows onto the plates of a capacitor when it is connected to a 12.0-V potential source. What is the minimum amount of work that must be done in charging this capacitor?

12×10⁻⁶ J.

Explanation:

The minimum amount of work that must be done in other to charge a capacitor = equal to the energy stored in a capacitor.

The Formula for the energy stored in a capacitor is given as,

E = 1/2QV Equation 1

Where E = Energy stored in a capacitor, Q = Charge on the capacitor, V = Potential difference connected across the plates of the capacitor.

Given: Q = 2×10⁻⁶ C, V = 12.0 V

Substitute into equation 1

E = 1/2(2×10⁻⁶)(12.0)

E = 12×10⁻⁶ J.

Hence the minimum amount of work that must be done in charging the capacitor = 12×10⁻⁶ J.

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