Question

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.)

Answer 1

Answer:

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