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Gauss' law, spherical symmetry

  • #1
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Homework Statement



Assume that a ball of charged particles has a uniformly distributed negative charge density except for a narrow radial tunnel through its center, from the surface on one side to the surface on the opposite side. Also assume that we can position a proton anywhere along the tunnel or outside the ball. Let FR be the magnitude of the electrostatic force on the proton when it is located at the ball's surface, at radius R. As a multiple of R, how far from the surface is there a point where the force magnitude is 0.79FR if we move the proton (a) away from the ball and (b) into the tunnel?

Homework Equations


F=qE=ma
E=σ/2ξo
a=σe/2εm

The Attempt at a Solution


I dont even know where to sstart
 

Answers and Replies

  • #2
BvU
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Hello tomorrow, welcome to PF :smile: !

Unfortunately, PF guidelines don't allow us to help if no effort at solution is shown. You can begin by picking (a) or (b) and write down what you do know, e.g. for the electrostatic field from a full sphere.

Since they tell you it's a narrow tunnel, I think you can safely disregard the effect from the tunnel on strength and direction of the field, so that's a giveaway for starters.

Do you realize this problem has a gravitational countertpart ? tunnel through the earth and a billiard ball, for example ?

What's a yay ?
 
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  • #3
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Hello tomorrow, welcome to PF :smile: !

Unfortunately, PF guidelines don't allow us to help if no effort at solution is shown. You can begin by picking (a) or (b) and write down what you do know, e.g. for the electrostatic field from a full sphere.

Since they tell you it's a narrow tunnel, I think you can safely disregard the effect from the tunnel on strength and direction of the field, so that's a giveaway for starters.

Do you realize this problem has a gravitational countertpart ? tunnel through the earth and a billiard ball, for example ?

What's a yay ?
Yay is a joyous exclaimation
if i start at part a) i get Fr=σq/4πεoR^2
then the ratio is F=0.79Fr
but how do i get r on its own?
 
  • #4
BvU
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Time to review the relevant equations, all variables and given/known data:
No idea what some of your variables stand for (you should list them in the problem statement), but if I guess:

F=qE=ma ##\qquad## F = qE is interesting. Apparently we need E
E=σ/2ξo ##\qquad##interesting too, but no business here. ##\sigma## is usually a surface charge density. No idea what ##\xi_0## stands for :rolleyes:
a=σe/2εm ##\qquad##idem dito. ##\epsilon## times ##m## or ##\epsilon_m## ? Who wants a anyway ?

(don't take this as offensive: I just want to show that one has good reasons to be as clear as possible)​


Leaves us with some curiosity wrt the "electric field: sphere of uniform charge" which of course you googled at the outset ?
(And up pops hyperphysics, and the word Gauss shows up too !)

We need ##\left | \vec E(r) \right | ## for ## r = R##, for some ##r_1 < R## and for some ##r_2 > R##

For ## r = R## you made a start. You get $$F_R={\sigma q \over 4π\epsilon_0 R^2}\ ,$$which does not have the dimension of a force. Can't be right.
what is ##\sigma## ? ##q## ? dimensions ?
Nor is $$F_R={\sum q \over 4π\epsilon_0 R^2}\ ,$$ which does not have the dimension of a force either.
But at least this has the dimension of an electric field strength ! The very ##\left | \vec E(R) \right | ## we were after !

So what is ##\left | \vec E(r) \right | ## for ## r ## if ##r < R## ?
And what is the expression for ##r > R## ?
 
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  • #5
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Hi, i did figure it out. thankyou. we do not have all the other variables it turned out to be a radial proportionality question, epsilon naught is an electric permittivity constant. in future i will be more specific.
thankyou for the break dow of the problem it did however increase my understanding of the subject
 

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