Why does a square of alternating charges have infinite neutral points?

[sivadas]

Homework Statement

four point charges +q,-q,+q,-q are placed at the four ends of a horizontal square of side 'a' .no of neutral points (where the electric field vanishes) is...

Homework Equations

The Attempt at a Solution

i think that there two dipole .. there is no null point due to dipole .but there to like charges ...due to that a neutral point exist in middle of square ...there fore i think that there exist only one neutral point...but ans provide in the key is infinity...how it will be..pls explain...this is a question from tifr 2009

 

[Qwertywerty]

i think that there two dipole .. there is no null point due to dipole .but there to like charges ...due to that a neutral point exist in middle of square ...there fore i think that there exist only one neutral point...but ans provide in the key is infinity...how it will be..pls explain...this is a question from tifr 2009

You have two dipoles placed in opposite direction . For any point d which is very far from these two dipoles (d >> distances between dipoles' monopoles) , the two will exert opposite electric field , equal in magnitude . Thus the net electric field at that point would be zero . This would be true for any far enough point and thus there would be ∞ null points , plus the center of course .

*I have assumed you know what the electric field is at any general point . Also I have not considered close points for I have already answered your question . I however think that electric field at a close by point would not be zero .

I hope this helps .

 

[BvU]

Hello sivadas, welcome to PF !

Is this a two-dimensional problem ?

PS I don't agree with qwerty²

 

[Qwertywerty]

PS I don't agree with qwerty2

My mistake ?

 

[BvU]

My mistake ?

Yes.

The fact that the field falls off faster than $1/r^2$ and also faster than $1/r^3$ doesn't mean it becomes zero at bigger distances: there is no $r<\infty$ for which $|{\bf \vec E}| < \epsilon$ for all $\epsilon > 0$. Except the r = 0 that was found already in the OP.

Sivadas' reasoning is correct - in the horizontal plane

 

[Qwertywerty]

The fact that the field falls off faster than 1/r21/r^2 and also faster than 1/r31/r^3 doesn't mean it becomes zero at bigger distances: there is no r<∞r 0. Except the r = 0 that was found already in the OP.

Have I said any of that ?

Suppose there is a dipole placed along the x-axis at the origin . Electric field at a point at a large distance along the y-axis is (magnitude) kp/r³ - where p is the dipole moment .

Now you place a similar dipole which is aligned in an opposite sense (direction) to the original at the same place as the first . Are you saying that the resultant of the two will not be zero at the far away on the y-axis (/or any other far away point for that matter) ?

 

[BvU]

Now you place a similar dipole which is aligned in an opposite sense (direction) to the original at the same place as the first

That is equivalent to removing the first dipole. I call that cheating . In the OP the charges are clearly not all in the origin.

 

[Qwertywerty]

That is equivalent to removing the first dipole. I call that cheating . In the OP the charges are clearly not all in the origin.

No , my point is this - if you take a far enough point , it won't matter whether the dipoles are in the exact same position or not - for a really far point , they will equivalently be at the same position , and then - what I said in my previous post .

 

[sivadas]

this is the actual problem- A17
look through that...

 

[Qwertywerty]

As BvU has said , it will be zero along all points on the z-axis , already giving ∞ points .

As I have said , it should be zero at far away points all over the plane and all over the three dimensional volume too - however , you need to use your discretion for my point , i.e. , it is upto you whether you consider Limit_d→∞ a/d = 0 .

You have two dipoles placed in opposite direction . For any point d which is very far from these two dipoles (d >> distances between dipoles and their monopoles) , the two will exert opposite electric field , equal in magnitude . Thus the net electric field at that point would be zero . This would be true for any far enough point and thus there would be ∞ null points , plus the center of course .