Respuesta :
Answer: a) E = 311,666.7 V/m, b) q = 1.54×10^-10c, c) F = 4.799×10^-5 N, d) 0.0629 J
Explanation:
Area = 700cm² = (700/100)² = 7² = 49m²
Distance between plates (d) = 0.3cm = 0.3/100 = 0.003 m
V = potential difference = 935v
A)
Recall that for a capacitor that V = Ed
Where E = strength of electric field.
935 = E× 0.003
E = 935/ 0.003
E = 311,666.7 V/m
B)
C =qV
Where q = magnitude of charge on capacitor.
We need to get the value for C before we can get for q.
C =ε0×A/d
Where ε0 =permittivity of free space = 8.85×10^-12
Hence we have that
C = 8.85×10^-12 × 49/ 0.003
C = 4.32×10^-10/0.003
C = 1.44×10^-7 F
But C=qV
1.44×10^-7 = q (935)
q = 1.54×10^-10c
C)
F=Eq
Where E = strength of electric field = 311,666.7 V/m, q = 1.54×10^-10c
F = 311,666.7 × 1.54×10^-10
F = 4.799×10^-5 N
D)
Total energy stored = cv²/2
Total energy = 1.44×10^-7 × 935²/2
Total energy = 0.0629 J
