Determine the energy stored in a parallel plate capacitor that is charged to a potential difference of 150 V. The plates are 4.50 mm apart and have an area of 87.5 cm2. Assume the space between the plates is filled with air.

E = 1.93×10^-7 J

Explanation: The formulae for getting the energy stored in a capacitor is given as

E =Cv²/2

Where E = energy stored in capacitor (joules), C = capacitance of capacitor (farad), v = applied voltage =150v.

The formulae implies that we need to get the capacitance of the capacitor (C) first before we can get the energy.

From the parameters given to us

Area of plates =A = 87.5cm² = 87.5/10000 = 0.00875m²

Distance between plates = d = 4.50mm = 4.50/1000 = 0.0045m

C = ε0A/d

Where ε0 = permittivity of free space = 8.85×10^-12 F/m

C = 8.85×10^-12 × 0.00875/ 0.0045

C = 7.743×10^-14/0.0045

C = 1.72×10^-11 F.

Recall that E = Cv²/2

E = ( 1.72×10^-11 × 150×150) /2

E = 3.87×10^-7/2

E = 1.93×10^-7 J

total distance travelled = 60 miles

total time taken =1.5 hours

average speed = total distance / total time

60 / 1.5

= 40

therefore, average speed = 40 m/h

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