(a) How much charge can be placed on a capacitor with air between the plates before it breaks down if the area of each plate is 9.00 cm2?
b) Find the maximum charge if polystyrene is used between the plates instead of air.
For a TeflonTM-filled parallel-plate capacitor having a plate area of 175 cm2 and insulation thickness of 0.0500 mm, determine the following:
(c) the capacitance
(d) the maximum voltage that can be applied to this capacitor in V

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge,    then the following gives us the potential difference,      the formula  may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get    using q = 8.00  × 10⁻¹⁹ c and  δv = 2.00 v, we get    the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.