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By a special feat of engineering, scientists manage to infuse space with a spherically symmetric volume charge density rho(r)= k/r^2 *e^(−ar), where k and a are constants.

a) What are the SI units of k and a?
b) What is the charge Q(r) contained within a sphere of radius r centered on the origin? (Hint: Q(r)6 is not equal to rho(r)*4/3 πr^3)
c) What is the total charge Qtot contained in all space?
d) Use Gauss’ law to calculate the electric field E(r) at all r >0.
e) Show that your result in part (d) satisfies ∇ · E = rho/apsolon0 at all r >0.

User Sir Robert
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Final answer:

The problem involves finding the SI units for k and a in a given charge density equation, calculating the charge within a sphere of radius r, the total charge in all space, the electric field using Gauss' law, and verifying that the result satisfies the divergence relation with the given charge density.

Step-by-step explanation:

The question involves determining the charge density, total charge, and electric field in a space where charge is distributed with a spherical symmetry according to a specific volume charge density function ρ(r) = k/r^2 * e^(−ar). Here are the steps to solving each part of the problem:

a) Units of k and a

To maintain dimensional consistency in the equation ρ(r) = k/r^2 * e^(−ar), the units of k must be coulombs per meter (C/m) since ρ has units of charge per volume (C/m^3), r has units of meters (m), and the exponential has no units. The constant a must have units of 1/m to make the argument of the exponential dimensionless.

b) Charge Q(r) within a sphere of radius r

To find Q(r), integrate the charge density over the volume of a sphere:
Q(r) = ∫ ρ(r') 4π r'^2 dr' from 0 to r.

c) Total charge Qtot

To find the total charge contained in all space, take the limit of Q(r) as r approaches infinity. This involves evaluating an improper integral.

d) Electric field E(r) using Gauss' law

Apply Gauss's law in integral form over a spherical surface of radius r. Use the symmetry of the problem to simplify the integral and solve for E(r).

e) Divergence of E

Show that the divergence of the electric field vector E(r) equals the charge density ρ divided by the vacuum permittivity ε0.

User Greg Rogers
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