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.

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards