Respuesta :
Answer:
Let's complete the question and so add the solution one at a time
First, the reminder of the question
Suppose that the charge density on either surface is ± 0.50×10−3 C/m2, the cell wall is 5.00 nm thick, and the cell-wall material is air.
Let's begin
A. Find the magnitude of~E in the wall between the two layers of charge.
Solution:
The electric field between two oppositely charged parallel plates is derived by summation of the fields produced by each plate.
Let's make the cell wall material to be referred to as air.
Eair=σ/0
=0.50×10-3/8.854×10-12(V/m)
= 5.647×10^7V/m
(b) Find the potential difference between the inside and the outside of the cell. Which is atthe higher potential?
Solution:
There is uniformity between the surfaces of the field.
Take, the potential difference between the inside and the outside of the cell to be
Vair=Eair*d,
Take,
Eair = electric field
d = the thickness of the cell wall.
Vair=Eair*d= 5.647×10^7V/m×5nm
= 0.282V
The positively charged outer surface of the cell wall produces a higher potential at the outside.
(c) A typical cell in the human body has a volume of 10-16m3. Estimate the total electric-field-energy stored in the wall of a cell this size. (Hint:Assume the cell is spherical, andcalculate the volume of the cell wall.)
Solution:
Take,
volume of the cell = radius.
Vcell =4/3πR^3cell
= 10-16m3
Let's find the Rcell
⇒Rcell =3(√3×10-16/4π)m
⇒Rcell= 2.879×10-6m
Therefore, the volume confined by the cell wall
= surface area * thickness
Volume of cell wall:
4πR^2cell*d
= 5.209×10-19m3
Then,
the total energy stored
U=1/2CV^2 = 12 A°/d (Eaird)^2
= 1/2 * Ad° E^2air = 1/2 4πR^2cell d° (σ/0)^2
= 1/2 * 4πR^2cell (σ^2/0)
=1/2 * 5.209×10-19 ×(0.50×10-3)^2/ 8.854×10-12J
= 7.354×10-15J
(d) In reality, the cell wall is made up, not of air, but of tissue with a dielectric constant of5.4. Repeat parts (a) and (b) in this case.
Solution:
In a dielectric, the permittivity ° is substituted by r°,
where,
r = dielectric constant.
The corrected value electric field as
Etissue = σ/r°
=5.647×10^7/5.4V/m
= 1.046×10^7V/m.
Following the electric field,
potential difference decreases by a factor of the dielectric constant.
Vtissue = Etissue*d = Eair*d/r = 0.282V/5.4
= 0.0522V
