Answer:
q=6.22*10^-10C
Explanation:
Two large metal plates of area 0.88 m2 face each other, 4.8 cm apart, with equal charge magnitudes but opposite signs. The field magnitude E between them (neglect fringing) is 80 N/C. Find |q|
E=α/∈, electric field within the plate
α=q/A
A=area of the plate
∈=is the permittivity
substituting , we have
The field magnitude E between them (neglect fringing)
E=q/A∈
q=EA∈
q=0.88*80*8.84*10^-12
q=6.22*10^-10C
Because light is being refracted
The atom number of an atom is the number of protons contained within its nucleus.
For the work, applicate formula:

According our data:
W = 12000 N * 1,5 m
W = 18000 J
The work done is <u>18000 Joules.</u>