Two parallel plates, each of area 3.37 cm2, are separated by 5.40 mm. The space between the plates is filled with air. A voltage of 8.55 V is applied between the plates. Calculate the magnitude of the electric field between the plates. Tries 0/20 Calculate the amount of the electric charge stored on each plate. Tries 0/20 Now distilled water is placed between the plates and the capacitor is charged up again to the same voltage as before. Calculate the magnitude of charge stored on each plate in this case. (Use κ = 83.0 for the dielectric constant of water.)

1, 1583.33 V/m

2, 4.72*10^-12 C

3, 39.2*10^-11 C

Explanation:

1

E = V / d

E = 8.55 / 5.4*10^-3

E = 8.55 / 0.0054

E = 1583.33 V/m

2

Capacitance, C = (k * e0 * A) / d, where k = 1

A = area of capacitor, 3.37 cm² = 3.37*10^-4 m²

d = plate separation, 5.4 mm

e0 = Constant, 8.85*10^-12

Applying these, we have

C = (1 * 8.85*10^-12 * 3.37*10^-4) / 5.4*10^-3

C = 29.82*10^-16 / 0.0054

C = 5.52*10^-13 F

Since Q = CV, then

Q = 5.52*10^-13 * 8.55

Q = 4.72*10^-12 C

3

We are given that k = 83, so

Capacitance, C = (k * e0 * A) / d

C = (83 * 8.85*10^-12 * 3.37*10^-4) / 5.4*10^-3

C = 2.475*10^-13 / 0.0054

C = 4.58*10^-11 F

Q = CV

Q = 4.58*10^-11 * 8.55

Q = 39.2*10^-11 C

no, they would

i'm a mechanic so i know all about this stuff.

none of the three examples you gave would do very good in a ice

motor oil gets extremely hot during operation and undergoes being thrown around, getting dirty,

syrups are sticky and thick. oil needs to be atleast a little runny to work in a car engine so that it does not clog parts like the camshaft or valves.

hope this !

a and c

explanation:

both a and c are correct answers to my knowledge, it's been a bit since i've done anything related to magnetic fields, but it should be right.