Dry air will break down if the electric field exceeds 3.0*10^6 v/m. what amount of charge can be placed on a parallel-plate capacitor if the area of each plate is 73 cm^2?

The charge on each plate of the capacitor is

with one plate positive and one negative, i.e.,

Solution:

According to the question:

Critical value of Electric field,

Area of each plate of capacitor,

Now, the amount of charge on the capacitor's plates can be calculated as:

Capacitance, C  =        (1)

Also, Capacitance, C =

And

Electric field, E =

So, from the above relations, we can write the eqn for charge, q as:

q =

q =

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