Evaluating an integral for an expanding, charged sphere

In summary, the charge Q is uniformly distributed over the sphere's volume, and the integral of the charge Q with respect to the center is equal to the charge Q multiplied by a factor of 3, if the sphere is smaller than the distance c from the center.
  • #1
ELB27
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Homework Statement


An expanding sphere, radius ##R(t) = vt## (##t>0##, constant ##v##) carries a charge ##Q##, uniformly distributed over its volume. Evaluate the integral [tex]Q_{eff} = \int \rho(\vec{r},t_r) d\tau [/tex] with respect to the center. (##t_r## is the retarded time and ##d\tau## is an infinitesimal volume element). Show that ##Q_{eff}≈Q(1-\frac{3v}{4c})##, if ##v<<c##.

Homework Equations


The definition of the retarded time: ##t_r ≡ t-\frac{d}{c}## where ##d## is the distance between observation point and charge.

The Attempt at a Solution


At any time ##t##, the volume charge density is constant: ##\rho(t) = \frac{Q}{\frac{4}{3}\pi (vt)^3}##. However, I believe that for the observer at the center, the density will not seem uniform because the retarded time depends on separation distance (and hence, signals emitted from points near the center will arrive simultaneously with earlier signals from far points). At a certain radius ##r<R##, the retarded time is ##t_r = t-\frac{r}{c}## and at this retarded time, the density at ##r## is ##rho(r,t_r) = \frac{Q}{\frac{4}{3}\pi (vt_r)^3} = \frac{3Q}{4\pi}\frac{c^3}{v^3}\frac{1}{(tc-r)^3}##. Integrating over the entire sphere at time ##t## and noting that the (spherical) ##\theta## and ##\phi## integrals together give ##4\pi##, [tex]Q_{eff} = 3Q\frac{c^3}{v^3}\int_0^{vt}\frac{r^2}{(tc-r)^3}dr = 3Q\frac{c^3}{v^3}\left[\ln\left(\frac{c}{c-v}\right) +\frac{3v^2-2cv}{2(c-v)^2}\right][/tex]
The problem is that if I expand the above expression in Taylor series about ##v=0##, I get ##Q_{eff}≈Q(1+\frac{9v}{4c})## in disagreement with the problem statement.
I have been thinking about this problem for quite some time and tried different approaches but none led to the correct approximation in the end and now I'm at a loss. Where is my argument wrong?
By the way, the author of my book argued before (when discussing potentials of a moving point charge) that the apparent volume in an integral such as the above is altered by the factor [tex] \frac{1}{1-\hat{d}\cdot\vec{v}/c} [/tex] where ##\hat{d}## is a unit vector pointing from source point to observation point. If so, the apparent volume (and hence, the effective charge) is bigger, not smaller as the approximation of the problem suggests.
Any comments/corrections will be greatly appreciated!
 
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  • #2
What is the limit on the integral?
 
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  • #3
DEvens said:
What is the limit on the integral?
Ah, I see... The largest radius from which the signal has just arrived at the time of evaluation: ##R = vt_r=c(t-t_r)##. From this, ##t_r=\frac{ct}{c+v}## and ##R=vt_r=\frac{vct}{v+c}## and this last expression constitutes the upper limit of the integral. With this, [tex]Q_{eff}=3Q\frac{c^3}{v^3}\int_0^{\frac{vct}{v+c}}\frac{r^2}{(tc-r)^3} dr = 3Q\frac{c^3}{v^3}\left[\ln\left(\frac{c+v}{c}\right)+\frac{v^2-2cv}{2c^2}\right][/tex] If ##v<<c##, this indeed gives ##Q_{eff}≈Q(1-\frac{3v}{4c})##.

Thank you very much! (Such a long OP and such a short answer :-p)
 
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1. What is the purpose of evaluating an integral for an expanding, charged sphere?

Evaluating an integral for an expanding, charged sphere allows us to determine the electric field or potential at any point outside the sphere. This is important in understanding the behavior and interactions of charged particles in the sphere's vicinity.

2. How is the integral for an expanding, charged sphere calculated?

The integral for an expanding, charged sphere is calculated using the Gauss's Law formula, which relates the electric flux through a closed surface to the enclosed charge. This involves setting up a Gaussian surface, determining the charge enclosed, and solving the integral using appropriate mathematical techniques.

3. What factors affect the value of the integral for an expanding, charged sphere?

The value of the integral for an expanding, charged sphere is affected by the charge distribution, the size of the sphere, and the distance from the sphere's center. It is also affected by the presence of other charged objects in the surrounding space.

4. Can the integral for an expanding, charged sphere be negative?

Yes, the integral for an expanding, charged sphere can be negative. This indicates that the electric field or potential at a certain point is directed in the opposite direction of the Gaussian surface normal. This can occur when the charge distribution is uneven or when there are other charged objects present in the vicinity.

5. How is the integral for an expanding, charged sphere used in practical applications?

The integral for an expanding, charged sphere is used in various practical applications, such as in the design of electronic circuits, in the analysis of electric fields in medical treatments, and in understanding the behavior of charged particles in space. It is also used in the development of technologies such as capacitors and antennas.

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