Suppose a undisturbed charge of magnitude +q at a certain place in the plane and then consider a point 'p' at a distance r from the charge. To calculate electric field at point 'p' we decide a gaussian surface. If I choose any random surface which include only point 'p' but not the charge q then ,according to gauss law then the enclosed charge is 0 then, the electric field will also be zero but, When I choose a surface which include both point 'p' and charge 'q' then we will get electric field.I know that second surface is correct but I want to know why choosing first surface leads to wrong answer?
$\begingroup$
$\endgroup$
6
-
2$\begingroup$ There is no "correct" or "incorrect" Gaussian surface; only useful or not-useful surfaces $\endgroup$– mike stoneCommented Sep 15, 2023 at 16:44
-
$\begingroup$ okk but why I get two different answers in choosing different surfaces. shuldn't electric field be same for all kind of surfaces $\endgroup$– ayuCommented Sep 15, 2023 at 16:54
-
$\begingroup$ You do not get different answers. You are misunderstanding how guassian surfaces are used. The useful surfaces are thiose such that symmetry guarentees that the flux through evergy point of the surface is the same. This is not the case for your surface about $p$. $\endgroup$– mike stoneCommented Sep 15, 2023 at 17:02
-
$\begingroup$ I did not understand your point. Can you elaborate it $\endgroup$– ayuCommented Sep 15, 2023 at 17:06
-
$\begingroup$ @ayu, Mike Stone is telling you that all Gaussian surfaces enclosing a charge have the same flux through them, regardless of their shape or where the charge is located relative to the surface. Because of this, you are free to choose the Gaussian surface that provides enough symmetry to make the calculation easy to solve. Examples: for a point charge, use a sphere with the charge located at its center; for an infinite line charge, use a cylinder with the line running down the length of the cylinder and through its center. $\endgroup$– David WhiteCommented Sep 15, 2023 at 19:29
|
Show 1 more comment
1 Answer
$\begingroup$
$\endgroup$
1
. . . . . according to gauss law then the enclosed charge is 0 then, the electric field will also be zero.
You are mixing up finding the electric field at a particular position and the electric flux passing through a surface.
A Gaussian surface which does not have a charge within it can have an electric field within it but the electric flux through the surface will be zero.
Electric field lines might help you visualise what is going on.
Suppose a volume which has an electric field within it but with no charges inside it is enclosed by a Gaussian surface.
In this case any electric field line which enters the volume (contributing to electric flux into the volume) also has to leave the volume (contributing to electric flux out).
Thus the net electric flux through the Gaussian surface is zero.
