Question about charge density in spherical coordinates! Please Help!?

Question

Let's suppose the electric field in some region is found to be E=k(r^3)r^-----> "r-hat", in spherical coordinates where k is a constant. How do you find the charge density, p, and the total charge containted in a sphere of radius R, centered at the origin?

Answer 1

You use Gauss's law, which states that the surface integral of the normal component of electric field intensity is equal to the volume integral of charge contained. (http://en.wikipedia.org/wiki/Gauss_law) You are given the electric field, and it only depends on r, not phi or theta. Therefore the surface integral for a sphere centered on the origin is just the surface area times E:

4*pi*r^2 * k*r^3 =4* pi*k*r^5

(I assume r-hat is the unit vector in the radial direction, so the field is always normal to the surface.)

The volume integral is rho*volume enclosed =rho* (4/3)*pi*r^3.

Equating the two

4*pi*k*r^5 = rho*(4/3)*pi*r^3

k*r^2 = (1/3)*rho

rho = 3*k*r^2 for any r. For r = R, rho = 3*k*R^2

EDIT: Forgot to include the permittivity e0. The surface integral is e0*E. Also, the link given is not right, it is http://en.wikipedia.org/wiki/Gauss%27_law

The correct answer is 3*e0*k*R^2

EDIT AGAIN (9/18): sparrowhawk is correct in that the differential form of Gauss' Law is easier to use when the electric field as a function of r is given and rho is sought. His answer is incorrect in that the div(E) in spherical coordinates is (1/r^2)d/dr[r^2*E(r)]. The result for your problem is then rho = ke0(1/r^2)*d/dr(r^5) = 5*k*e0*r^2. My error in applying the integral form was not realizing the charge density is not uniform, therefore the volume integral of charge density is not rho*(4/3)*pi*r^3. You can still use the integral form: 4*pi^r^2*e0*E = integral[rho(r)]dV. Since rho depends only on r, this gives [for E=k*r^3)] 4*pi*e0*k*r^5 = integral[rho(r)*4*pi^r^2]dr. Take the derivative of both sides to get 4*pi*e0*k*5*r^4 = 4*pi*r^2*rho(r). This gives rho(r) = 5*k*e0*r^2, the same as the result from using the divergence.

The total charge contained can be obtained by integ[rho(r)*4*pi*r^2]dr, or by the surface integral[e0*E(r)*]dS. Remember dS in sperical coordinates is r^2*sin(theta)dThetadPhi. If there is no variation in phi or theta, this becomes dS=4*pi*r^2, so the surface integral of e0*E(r) is just 4*pi*r^2*e0*E(r). Thus the total contained charge in your problem is 4*pi*e0*k*r^5

Answer 2

I'd probably go for the possibly inexact solution of dividing the drop into two hemispheres and look at the strain on the surface caused by the charge. If that exceeds the surface tension, than I know it is unstable. This would be the same as calculating the max pressure rating of a spherical tank. ******** I remember very, vaguely doing it Zo Maar's way, so I am sure he is right, but I can at least ballpark it pretty good. Here goes: Voltage inside the droplet is: V= kQ/r The surface area of the droplet is: 4 π r² Density of the charge on the droplet: Q/4 π r² The cross-sectional area of the sphere is: π r² The circumference of the sphere is: 2 π r ......... Now calculate the force pushing the two halve of the droplet apart: Force = k * Q/r² * Charge Density * Cross-sectional area = k Q/r² * Q/4 π r² * π r² = kQ²/4 r² Force/unit length = surface tension σ = kQ²/4 r² * 1 /( 2 π r) σ =kQ²/8 r³π therefore Q = √ 8 r³π σ /k Pretty good for an estimate. ********* What I attempted to calculate was the charge where the drop would literally tear itself apart. Zo Maar's analysis should be where the drop is unstable. In other words, when it starts to stretch, it will keep going. My number was 1/sqrt 2 less when it should be more, which means I probably have an error someplace. Nice question.

Answer 3

div(E)=p/e0

to find the charge density all you have to do is take the divergence of E.

In case the divergence is equal to (1/r^2*d/dr(r^2*dE/dr))
or
15*k*r^4=p/eo k=1/4pieo

To find the total in closed charge use the integral form of Gauss' law

or E(R)*a=Q inside/e0

Answer 4

Al tough, this won't directly answer your question, you can find resources available at the link I provided you below.
Good Luck
Fb