Purcell 1.77 Electron jelly

  • Thread starter Thread starter Rob2024
  • Start date Start date
AI Thread Summary
The discussion revolves around the challenges of minimizing the energy method for calculating potential energy in the context of a charged sphere. The initial approach using energy equations was deemed ineffective, leading to the realization that the force balance method yields correct results. There is confusion regarding the symbols used, particularly the sphere's radius, which is referred to inconsistently. A suggestion is made to reevaluate the potential due to the negative charge and to reference established formulas for potential inside a uniformly charged sphere. Ultimately, the user acknowledges a misapplication of the potential energy concept, which has been clarified through the discussion.
Rob2024
Messages
37
Reaction score
6
Homework Statement
Imagine a sphere of radius a filled with negative charge of uniform
density, the total charge being equivalent to that of two electrons.
Imbed in this jelly of negative charge two protons, and assume that,
in spite of their presence, the negative charge distribution remains
uniform. Where must the protons be located so that the force on
each of them is zero? (This is a surprisingly realistic caricature of
a hydrogen molecule; the magic that keeps the electron cloud in
the molecule from collapsing around the protons is explained by
quantum mechanics!)
Relevant Equations
##U= \frac{k Qq}{r}, E = \frac{kQ}{r^2}##
I initially tried energy method, I realized this cannot be minimized.

##U = \frac{kq^2}{2r} - \frac{2 k qq'}{r} , q' = \frac{2e r^3}{R^3},q = e ##
##U = \frac{kq^2}{2r} - \frac{2 k q\frac{2e r^3}{R^3}}{r} ##
##= k q^2 ( 1/2r -4 r^2/R^3) ##
##= \frac{1}{2} k q^2 ( r - 8 r^2/R^3) ##
##U' \sim -1/2r^2 - 8 r/R^3 = 0 ##

The force balance method works correctly.

##E = \frac{k q}{4 r^2}, E' = \frac{k q'}{r^2} = \frac{ k 2 q r^3/R^3}{r^2} ##
##r = \frac{1}{2} R##


I cannot figure out what is causing this inconsistency.
 
Physics news on Phys.org
The meanings of your symbols are not always clear and you have not provided any explanations of the rationale. But I think I can guess. Also you have used '##R##' for the sphere's radius but the question calls it 'a'.

Consider the potential a distance ##r## from the centre of the sphere due to the negative charge alone. The potential is not ## \frac{k \frac{-2e r^3}{R^3}}{r}##.

You can try working out the correct expression for yourself or do a search such as 'potential inside a uniformly charged sphere'.
 
  • Like
Likes berkeman, Rob2024 and TSny
Thanks, this worked. I used the potential energy incorrectly.
 
  • Like
Likes TSny, Steve4Physics and berkeman
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top