B Static Point Charge Should Have Zero Effect

[bryanso]

TL;DR Summary

A static point charge has field lines radiating in all directions, don't they cancel each other?

How can we detect electrical effect of a static point charge at all?

I think of a point charge like a sea urchin. With field lines going outwards in all directions (for +ve). So the vector pointing at me directly should be canceled perfectly by the vector going away from me. And so each line pointing at any one direction would be canceled by the one going in the opposite direction. Isn’t field lines simply vectors that superimposed and add up? How can a static charge have any effect at all??

If physicists use vector to represent field lines of a static point charge, don't they have to honor all vector addition laws and admit no effect can be detected anywhere surrounding the charge?

 

[berkeman]

How can we detect electrical effect of a static point charge at all?

Use a test charge?

 

[Frabjous]

Using that logic you would not feel the wind either.

 

[bryanso]

Using that logic you would not feel the wind either.

Not really. Wind doesn't radiate in all direction like a sea urchin. But point charge does.

 

[berkeman]

Not really. Wind doesn't radiate in all direction like a sea urchin. But point charge does.

So what? At the point in space that you place a test charge, there is only one vector sum of the E fields from your stationary charges. Same for the wind -- if you place a tiny anemometer at a point in space, there is only one vector sum of the wind force at that point, assuming stationary sources for the wind.

 

[bryanso]

Use a test charge?

Hmm... not sure how to counter your argument.

 

[berkeman]

BTW, the wind analogy is not perfect. Gusts of wind coming from different sources do interact with each other, while E-fields coming from different sources do not interact; they just add vectorally.

 

[bryanso]

-q ... q (sorry can't get spacing to work)
<-------- Q --------> test charge here

Q is a point charge. q is the electric field vector pointing to the right.

At test charge position, there should be no electric field. Vector q is canceled by vector -q.

 

[Frabjous]

That is 2 point charges, not one. The small amount of displacement creates a dipole.

 

[PeroK]

-q ... q (sorry can't get spacing to work)
<-------- Q --------> test charge here

Q is a point charge. q is the electric field vector pointing to the right.

At test charge position, there should be no electric field. Vector q is canceled by vector -q.

If you throw one sea urchin to the left and another sea urchin to the right with the same speed, then the momentum of your sea urchins cancels out and the total momentum is zero. But, if someone gets hit by one of the sea urchins then they'll know about it!

 

[topsquark]

-q ... q (sorry can't get spacing to work)
<-------- Q --------> test charge here

Q is a point charge. q is the electric field vector pointing to the right.

At test charge position, there should be no electric field. Vector q is canceled by vector -q.

Say you have a positive point charge at the origin. Put yourself at the point (5,0). The electric field points away from a positive charge so you will detect a component of the electric field from the charge pointing in the +x direction. What you are saying is that you should also feel one from the opposite direction. But how are you going to feel that? You are on the wrong side of the charge to do that.

When you are in an electric field due to a source you can only measure the field from one direction... the direction from the source. You can't measure the field from a position opposite of where you are. The field doesn't point in that direction at your position.

-Dan

 

[malawi_glenn]

Seems to me that you have missunderstood what electric field lines are. They are a "map" of the electric force exerted on a positive test charge by a source charge.

Black: positive source charge. Red: test charge

and so on.

In order to "map" the magnitude of the force, we use density. The denser the electric field lines, the larger the force.

So, this what a "force map" (electric field lines) from a positive source charge looks like.

In order to find the direction of the electric force on your test charge at a given point, you use the direction of that field line.
In order to find the magnitude, you must have some reference value.

What will the electric force be at this point then?

Well the direction is simple to find. What about the magnitude?
This point is located where the field lines are denser compared to the previous example. This means stronger force.
Maybe something like this

 

[bryanso]

OK OK. Now I understand. They are not pure mathematical vectors that just cancel out BEFORE acting on other things.

More like two bullets firing, one to the left, one to the right. The vectors cancel out. But people die! (Perok... thanks for your sea urchin throwing thought experiment...)

 

[DaveE]

I still think you are misunderstanding what the field is/means. Think of it as a map of the force a test charge would experience at each point in space. But the only part of the field that matters to a test charge is the value at its location. The rest of the field is a description of "what if" the test charge was in a different place.

(Yet) another analogy: Think of it like a topographic map that shows you your altitude at each point in a mountain range. Yes, you have data about what the altitude is "over there", but that doesn't affect you. You only feel the value at your location.

 

[Orodruin]

They are not pure mathematical vectors that just cancel out BEFORE acting on other things.

They are not vectors at all. They are the flow lines of the electric field (which assigns one vector to each point in space).

A flow line of a vector field is a curve that is everywhere parallel to the field.

(Yet) another analogy: Think of it like a topographic map that shows you your altitude at each point in a mountain range. Yes, you have data about what the altitude is "over there", but that doesn't affect you. You only feel the value at your location.

Coincidentally, that would correspond to a dual vector field rather than a tangent one (which is what flow lines are related to) but the general idea is the same.

 

[bryanso]

Ok ok. Much better. Thx

 

[SammyS]

Hmm... not sure how to counter your argument.

It's very difficult to counter a valid argument.

 

[phinds]

It's very difficult to counter a valid argument.

Nah. Forum rules prohibit specifics, but we have tens of millions of people in this country that do it all the time with the greatest of ease.

 

[berkeman]

It looks like the OP's question has been answered to their satisfaction, so this is a good place to tie off this thread. Thanks everybody.