To get an idea how big a farad is, suppose you want to make a 1.0-F air-filled parallel-plate capacitor for a circuit you are building. To make it a reasonable size, suppose you limit the plate area to 1.4 cm2 . Part A
What would the gap have to be between the plates?
Express your answer using two significant figures and include the appropriate units.
Part B
Is this practically achievable?
Is this practically achievable?
a. yes
b. no

The gap between the plates will be 1.2 x 10^-15 m

No, this is not practically achievable.

Explanation:

Capacitance = 1.0 F

area of plate = 1.4 cm^2 = 1.4/10000 m^2 =  m^2

distance = ?

We use the equation

=  *ε

C is the capacitance

where A is the area

d is the distance of separation of plates

ε is the permeability of free space = 8.854×10^-12 F⋅m−1

substituting values, we have

1 = * 8.854×10^-12

distance between plates = 1.2 x 10^-15 m

This is not practically achievable in real life

c

explanation:

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day