Electrostatics, Neutral Points

  • #1
59
4
Untitled.png

Statement
Four point charges, Q, -Q, Q, -Q, are placed at the four ends of a horizontal square of side a as shown in the figure above. What are the neutral points?


The attempt at a solution
The square is not exactly aligned parallel in its plane, say XY.
So, the center is not a neutral point.

I can not calculate the dipole moments. I would really appreciate any hint.
Thanks.
 

Attachments

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
2019 Award
13,588
3,283
Hello VSayantan, :welcome:
You erased the second item: relevant equations ! PF culture wants you to provide them -- it also helps us help you

So what equation describes a 'neutral point' ? (And, for that matter: what exactly is a neutral point ?)
I can not calculate the dipole moments
Why not ? Isn't that something like ##\sum q_i\vec r_i## ?
 
  • #3
59
4
Hello VSayantan, :welcome:
You erased the second item: relevant equations ! PF culture wants you to provide them -- it also helps us help you

So what equation describes a 'neutral point' ? (And, for that matter: what exactly is a neutral point ?)
Why not ? Isn't that something like ##\sum q_i\vec r_i## ?

Thanks BvU.
Sorry I did not know that!

A neutral point is one where the electric field due to all the contributing charges cancel out.
So, if the square were placed parallel in its plane, the field at the center would have been zero. But it is not. Which I think strains the configuration giving a dipole moment.

I know expression for dipole moments, but I don't know what the coordinates of the two lower charges.
Do you think the orientations of the axes do not matter in this case?
 
  • #4
BvU
Science Advisor
Homework Helper
2019 Award
13,588
3,283
A neutral point is one where the electric field of all the contributing charges cancel out.
Fine with me. The reason I asked is that for some folks neutral means V = 0. But in your case you want ##\vec\nabla V = 0##, right ?

So, the center is not a neutral point.
What makes you write that ? With your definition I would claim it is !​

So you can either write the four contributions to ##\vec E## and require the sum to be zero, or write the four contributions to ##V## and require the derivatives to be zero.

I don't know what the coordinates of the two lower charges
If you know them for the upper two, you also know them for the lower two !?


However, the more I look at the wording of the problem statement
a horizontal square
, the more I fear that the picture is a poor man's 3D attempt and the charges are simply located at $$(a/2,a/2,0)\quad (a/2,-a/2,0)\quad (-a/2,-a/2,0)\quad (-a/2,a/2,0)$$what do you think of that ?

.
 
  • #5
59
4
Fine with me. The reason I asked is that for some folks neutral means V = 0. But in your case you want ##\vec\nabla V = 0##, right?

yes.
I want to find the points where ##\vec E=-\vec\nabla V = 0##

What makes you write that ? With your definition I would claim it is !.
So you can either write the four contributions to ##\vec E## and require the sum to be zero, or write the four contributions to ##V## and require the derivatives to be zero.

Ok.


However, the more I look at the wording of the problem statement , the more I fear that the picture is a poor man's 3D attempt and the charges are simply located at $$(a/2,a/2,0)\quad (a/2,-a/2,0)\quad (-a/2,-a/2,0)\quad (-a/2,a/2,0)$$what do you think of that ?
So, if I understand correct it is a simple square, sitting nicely on the XY-plane.

But, the potential at a point z above the plane is
##V(z)=\frac 1 {4\pi\epsilon_0} [\frac Q b - \frac Q b + \frac Q b - \frac Q b]##,
where ##b=\sqrt(z^2 + \frac {a^2} 2)##

Which is zero.
There is another like point below the plane, where the potential is zero.
But these two points give ##\vec E=constant##

Thus there is no other neutral point apart from the center. Am I right?
 
Last edited:
  • #6
BvU
Science Advisor
Homework Helper
2019 Award
13,588
3,283
Hehe, you typed your reply within the [quote] ... [/quote] tags. never mind.

I agree that ##V(0,0,z)= 0\ .\ \ ## I don't see why that would give ##\vec E= {\rm constant} \ne 0\ ##.
 
  • #7
59
4
Oops.
##V=0##
will also give ##\frac {dV} {dz}=0##
Of course!
So, there are three neutral points for this configuration.

Hmm.
But the question does not provide such an option!

Thanks @BvU a lot, for helping.
 
  • #8
BvU
Science Advisor
Homework Helper
2019 Award
13,588
3,283
Not done yet. ## V(0,0,z)= 0 \ ## for all ##z## only leads to the conclusion that the ##z## component of ##\vec E ## is zero on the entire ##z## axis (not just 3 points but a whole line !).

ToDo: show that the ##x## and ##y## components are zero there too
 
  • #9
59
4
Not done yet. ## V(0,0,z)= 0 \ ## for all ##z## only leads to the conclusion that the ##z## component of ##\vec E ## is zero on the entire ##z## axis (not just 3 points but a whole line !).

ToDo: show that the ##x## and ##y## components are zero there too
Along the ##z##-axis there are no other components i.e., ##E_x = 0## and ##E_y =0##.
So, ##\vec E ## is zero for all points on the ##z##-axis!
 
  • #10
BvU
Science Advisor
Homework Helper
2019 Award
13,588
3,283
there are no other components
In the sense that individual contributions from the four charges cancel, I agree. Well done !

(You picked up on ##\ \LaTeX\ ## too ! very good !)
 

Related Threads on Electrostatics, Neutral Points

  • Last Post
Replies
14
Views
5K
  • Last Post
Replies
12
Views
3K
Replies
23
Views
2K
  • Last Post
Replies
5
Views
4K
Replies
4
Views
2K
Replies
1
Views
4K
Replies
4
Views
8K
Replies
2
Views
8K
  • Last Post
Replies
7
Views
12K
Replies
3
Views
1K
Top