A parallel-plate capacitor is made from two aluminum-foil sheets, each 3.0 cm wide and 5.00 m long. Between the sheets is a mica strip of the same width and length that is 0.0225 mm thick. What is the maximum charge?

This question is incomplete, the complete question is;

A parallel-plate capacitor is made from two aluminum-foil sheets, each 3.0 cm wide and 5.00 m long. Between the sheets is a mica strip of the same width and length that is 0.0225 mm thick. What is the maximum charge?

(The dielectric constant of mica is 5.4, and its dielectric strength is 1.00×10⁸ V/m)

the maximum charge q is 716.85 μF

Explanation:

Given data;

with = 3.0 cm = 0.03

breathe = 5.0 m

Area = 0.03 × 5 = 0.15 m²

dielectric strength E = 1.00 × 10⁸

∈₀ = 8.85 × 10⁻¹²

constant K = 5.4

maximum charge = ?

the capacitor C = KA∈₀ / d

q = cv so c = q/v

now

q/v = KA∈₀ / d

q =  vKA∈₀/d = EKA∈₀

we substitute

q = (1.00 × 10⁸) × 5.4 × 0.15 ×  8.85 × 10⁻¹²    

q = 716.85 × 10⁻⁶ F

q = 716.85 μF

the maximum charge q is 716.85 μF

when jerome paddle in flowing river towards east the net speed of jerome will increase

so the net speed of jerome will be given as

now the speed of paul will be same as it is given as he is running on the ground

so

so here we can say since the speed of jerome is more so he will win the race

now as we know the distance so the time difference is given as

so jerome will reach first and the time difference will be 203.55 s or 3.4 min