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# Application of Gauss' law

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Consider a spherically symmetric configuration: A spherical charge distribution has radius R_A and *nonuniform* charge density rho(r) = rho_zero R_A / r. It is encircled by a conducting shell with inner radius R_B and outer radius R_C (R_A < R_B < R_C). Therefore, there are four regions:
I. charged ball: 0 < r < R_A
II. gap (air) R_A < r < R_B
III. conductor R_B < r < R_C
IV. outside (air) r > R_C

a) Find the electric field in all four regions as a function of radius (E_I(r), E_II(r), E_III(r), and E_IV(r))

b) Find the electric potential in all four regions as a function of radius (V_IV(r), etc.) Take V=0 at infinite radius.

Note : R_A means R subscript A etc.

https://brainmass.com/physics/electric-power/application-gauss-law-73463

#### Solution Preview

SOLUTION

Fundamentals in a nutshell :

1. Gauss' law : The surface integral of the normal component of electric field intensity vector (E) over a closed surface (S)of any shape is equal to the total charge enclosed (Q) divided by the permittivity of free space (Є0).

[The surface integral of the normal component of electric field intensity vector over a closed surface is also known as flux through the closed surface]

In the form of equation we can state Gauss' law as : ∫E.ds = Q/Є0
S

2. Electric field intensity E at a point is related to the potential V at that point be the relation
E = - dV/dr (electric field intensity at a point is equal to the negative of the potential gradient at that point.

Or V = - ∫Edr

RC RB

RA

r+dr r I II III IV

Charge density ρ(r) = ρ0RA/r

Calculation of charge on the inner sphere

Let us consider two imaginary spherical shells with radii r and r+dr co-centric with the charged sphere (shown dotted in the fig.). Let us derive an expression for the charge contained with in the imaginary sphere of radius r.

The charge density at radius r = ρ(r) = ρ0RA/r .....(1)

Charge contained with in the shell of thickness dr (between the two imaginary spheres) = dQ = Charge density [ρ(r)] x Volume of the shell .......(2)

Volume of the shell = Volume of the outer imaginary sphere - Volume of inner imaginary sphere = 4/3 Π (r+dr)3 - 4/3 Π r3 = 4/3 Π [r3 + rdr2 + 2r2dr + r2dr + ...

#### Solution Summary

This is a good example of application of Gauss' law to determine the electric field and potential in different regions.

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