Solving Gauss's Law Problem: Two +q Charges Separated by d

[Avinash Verma]

Consider two point charges +q,+q. Separated by distance d.
now there exist a point P on the line joining these two charges where electric field cancels out at distance d/2 from the charge.
If we make a Gaussian surface at this point and work out the surface integral it won't be zero.
since two field come in at this point to cancel each other.
My argument for why those field line coming in at point P has to meet is symmetry they cannot go right or left doesn't make any sense.
But there is no charge there!
how is it possible?
or is there any mistake in my question .
please tell your thoughts.

 

[rude man]

Depends on the Gaussian surface you have chosen.

If the surface excludes either charge, the surface integral of E (or D) is zero. If both charges ar included within the surface the surface integral of D = 2q. Etc.
Not sure what your question really is.

 

[BvU]

work out the surface integral it won't be zero

It will be, as rudy says. To prove otherwise, show us and do the math !

 

[Avinash Verma]

Depends on the Gaussian surface you have chosen.

If the surface excludes either charge, the surface integral of E (or D) is zero. If both charges ar included within the surface the surface integral of D = 2q. Etc.
Not sure what your question really is.

the image shows the gaussian surface around which integral need to be taken and am talking just about that single field line (along the line joining the two charges )has to come in and cancel each other because of the symmetry and the surface in the image has lines coming in.
we have a point where the field is zero if we go a little to the left (along the line joining the two charges)than we have a field coming same goes for the right side.
Flux through a closed surface without any charge inside.
(I may be wrong some where but where?)

 

[BvU]

Do the math. Sphere with radius $>0$ will have field lines coming in from left & right, but going out top & bottom

 

[Orodruin]

Sorry, but you are just simply wrong. You cannot compute surface integrals by looking at field lines - that is just a heuristic. If you want to draw the lone you are talking about, then you should also be drawing the vertical line in the middle of the charges. This is where the lines you are talking about ”come out” if you wsnt to use the heuristic.

 

[vela]

the image shows the gaussian surface around which integral need to be taken and am talking just about that single field line (along the line joining the two charges )has to come in and cancel each other because of the symmetry and the surface in the image has lines coming in. we have a point where the field is zero if we go a little to the left (along the line joining the two charges)than we have a field coming same goes for the right side. Flux through a closed surface without any charge inside. (I may be wrong some where but where?)

Since there is cylindrical symmetry in this problem, consider a cylindrical Gaussian surface. You're only looking at the flux through the two ends of the cylinder. What about the flux through the cylindrical part of the surface?

 

[BvU]

View attachment 228858

My argument for why those field line coming in at point P has to meet is symmetry they cannot go right or left doesn't make any sense.
But there is no charge there!
how is it possible?
or is there any mistake in my question .
please tell your thoughts.

You shy away from doing the math, it seems.
Well, here is another qualitative argument: the blue field lines bend off to infinity. You draw two black field lines that stop. That is unphysical(*). They too bend off to infinity, either left or right.

(*) Unless they are smack on the axis, in which case the contributing area on the gauss surface is zero.

A test charge would not come to a stop, but shoot away. The origin is an unstable saddle point.