Question

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?

Answer 1

Answer:

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.

Answer 2

The minimum amount of work that must be done in charging this capacitor will be "12 × 10⁻⁶ J".

Potential difference and Energy

According to the question,

Charge on the capacitor, Q = 2.00 μC or,

                                              = 2 × 10⁻⁶ C

Potential difference, V = 12.0 V or,

                                     

We know the relation of energy stored,

→ E = $\frac{1}{2}$ QV

By substituting the values,

      = $\frac{1}{2}$ × (2 × 10⁻⁶) (12.0)

      = 12 × 10⁻⁶ J

Thus the response above is correct.

Find out more information about work done here:

https://brainly.com/question/87152