Mass, charge etc. at a single point

  • Thread starter Thread starter ForMyThunder
  • Start date Start date
  • Tags Tags
    Charge Mass Point
AI Thread Summary
Modeling a force as directed to a single point is feasible under specific conditions, such as when dealing with spherically symmetric mass or charge distributions. For gravitational forces, the Earth can be treated as a point mass at its center due to its homogeneity and symmetry. In contrast, electric charges, which can have different signs, require careful consideration; for example, a proton and an electron at a distance do not cancel out their electric fields. The concept of multipole expansion is used to account for charges at large distances. Understanding these principles is crucial for accurate calculations in physics.
ForMyThunder
Messages
149
Reaction score
0
I have come across a question. When is it possible to model a given force as being directed to a single point? I know that if you are outside of the Earth, you can say that the mass of the Earth is concentrated at its center. But if you had, say, three equidistant points of charge, you can model them as a single point (I tried doing this while calculating the work it would take to move an electron from a finite distance to an infinite distance and the answer I got was wrong). What is the general principle as to doint this?

I figure it has something to do with the assumption that the Earth is homogeneous.
 
Physics news on Phys.org
You can't model charge in the same way because charge can have different signs. For instance if we have proton and an electron and we look at it far away you might be tempted to say that the electric field is zero because the charges cancel out, however there is a field and we can calculate it.

There's a general way to account for charges at large distances it's called a multipole expansion.

EDIT: Here's a good introduction -- http://www.pa.msu.edu/~duxbury/courses/phy481/Fall2009/Lecture14.pdf
 
I figure it has something to do with the assumption that the Earth is homogeneous.
It does not have to be homogeneous, but it has to be spherically symmetric. It's an important theorem in Newtonian mechanics that the external gravitational field of a spherically symmetric mass distribution is the same as that of a point mass located at its center. A similar theorem holds also for spherically symmetric charge distributions.
 
Susskind (in The Theoretical Minimum, volume 1, pages 203-205) writes the Lagrangian for the magnetic field as ##L=\frac m 2(\dot x^2+\dot y^2 + \dot z^2)+ \frac e c (\dot x A_x +\dot y A_y +\dot z A_z)## and then calculates ##\dot p_x =ma_x + \frac e c \frac d {dt} A_x=ma_x + \frac e c(\frac {\partial A_x} {\partial x}\dot x + \frac {\partial A_x} {\partial y}\dot y + \frac {\partial A_x} {\partial z}\dot z)##. I have problems with the last step. I might have written ##\frac {dA_x} {dt}...
Thread 'Griffith, Electrodynamics, 4th Edition, Example 4.8. (Second part)'
I am reading the Griffith, Electrodynamics book, 4th edition, Example 4.8. I want to understand some issues more correctly. It's a little bit difficult to understand now. > Example 4.8. Suppose the entire region below the plane ##z=0## in Fig. 4.28 is filled with uniform linear dielectric material of susceptibility ##\chi_e##. Calculate the force on a point charge ##q## situated a distance ##d## above the origin. In the page 196, in the first paragraph, the author argues as follows ...
Back
Top