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Alex Ar [27]
3 years ago
13

(An easy problem which will be graded). Later in the quarter we will spend some time solving the diffusion equation Op(r, t) = D

V4p(r, t), at where D is called the diffusion constant. For now, let us consider a solution which represents a steady state solution (i.e apat = 0) and which has spherical symmetry. In that case the above reduces to 1d (20p) = 0 p2 dr dr / a) Obtain the solution of this second-order differential equation. It will have two arbitrary constants. b) Application: diffusion from a large distance to a spherical absorber. Consider p(r) to represent the density of some particles which are diffusing through some media. Think of them as food for a cell. Their density at large distance (infinity) is po. There is a spherical absorber of radius R centered at the origin, which means p(R) = 0. (This absorber represents the cell.) Show that the particular solution that satisfies these two condition is p(r) = po(1 – R/r). An interesting result: The current density (the number of particles passing through unit area per unit time) is given by j(r) = -Ddp/dr = -DpoR/r2. (The minus sign means that the flow is inward.) The magnitude of the current at a distance r is I = 47r2|j(r)= 47 DPOR. Note that this result says that the rate at which particles are absorbed is proportional to the radius of the sphere, not to the area of the sphere as would be the case if we were dealing with light being absorbed. Of course light does not travel by diffusion!

Physics
1 answer:
alexandr402 [8]3 years ago
7 0

Answer:

Explanation:

Answer is in the attachment below:

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an airplane is flying through a thundercloud at a height of 2000m (this is very dangerous thing to do because of updrafts, turbu
Vedmedyk [2.9K]

Answer

In this question we have given,

Height of plane, h1=2000m

Height at which charge concentration is 40C, h2=3000m

Height at which charge concentration is -40C, h3=1000m

charge concentaration, q1=40C

charge concentaration, q2=-40C

let the charge concentrations at height h2 and h3 as point charges

Now we will first find the electric feild on plane due to positive charge q1=40

E1= k*q1/(h1-h2)..............(1)

Here k=8.98755*10^9N.m^2/C^2

q1=40C

put values of k, q1 , h1 and h2 in equation 1


[tex]E1=(8.98755*10^9)*(40)/(2000-3000)^2\\

E1=[tex]E= 359502+359502\\E=719004 V/mV/m[/tex]

similarly electric feild due to negative charge q2=-40

[tex]E2=(8.98755*10^9)*(-40)/(2000-1000)^2\\

E2=359502V/m

Total electric feild E at the aircraft is given as

E= E1+ E2\\...............(2)

Put values of  E1 and E2 in equation2

\\E=359502+359502\\E= 719004V/m\\

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Where λ is the wavelength of the light; 1282.17 nm = 1282.17×10^-9 m
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Hence 1/(1282.17 ×10^-9) = 1.09737× 10^7 [1/n1^2 – 1/25^2]
Some rearranging and collecting up terms:
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