-1
$\begingroup$

In Gauss Law for Electrostatics, Electric flux term contains Electric field due to charges that are both inside and outside the Gaussian surface. If we use Gauss Law to calculate electric field at a point near the infinite line of charge we make Gaussian cylinder (as Gaussian surface) coaxial with the line of charge in such a way that point is on the cylindrical surface. Now if we place another infinite line of charge outside our Gaussian surface. Will the electric field computed by using our previous Gaussian cylinder gives resultant electric field as Gauss's law contains electric field due to both the charges that are inside and outside the Gaussian Cylinder. Kindly Explain it I am very much confused...

$\endgroup$
1

2 Answers 2

1
$\begingroup$

Gauss' law (in integral form) gives you the total flux over the entire surface. If the situation is symmetrical, and the surface is chosen appropriately such that the electric field is constant over the whole surface, then it's easy to find the electric field: you just divide the flux by the area of the surface.

This is true for the one-wire case - by symmetry, a cylinder surrounding a line of charge is going to have the same electric field everywhere, so the total flux and electric field at a point are proportional. But for the two-wire case, this symmetry is broken: in one privileged direction there's a whole other line charge! This means that you can't straightforwardly find the electric field. You still know the total flux over your Gaussian cylinder - that depends only on the charge enclosed - but the relationship between the flux and the electric field is more complicated.

(In the case of two line charges, the solution still ends up being quite simple. You can just find the electric field from each line charge separately (completely ignoring the other one), and then add them together, and by the principle of superpositon you will have found the solution.)

$\endgroup$
2
  • $\begingroup$ But if we have a gaussian surface and a charge out of this surface how will the flux from the external charge be zero if the electric field will drop proportional to r^2 ? $\endgroup$ Commented May 16, 2019 at 23:27
  • $\begingroup$ @ado sar The field lines pass through the gaussian surface twice: once from the outside as they enter the volume enclosed, and once from the inside as they leave the volume enclosed. Because these are coming from different directions, they have opposite signs, and when you integrate over the whole surface they all cancel out. $\endgroup$
    – Tetra
    Commented May 18, 2019 at 2:03
1
$\begingroup$

Gauss's law does not tell us the field, it tells us the total flux through the surface i.e. the integral of the field over the surface. For charges outside the surface there will some field lines that start at the charge, enter the surface, pass through the space inside surface and then exit the surface again. These field lines make a zero contribution to the total flux since their entry and exit cancel out to zero. The only field lines that make a contribution to the total flux are field lines that enter the surface and end inside it.

Since field lines can only begin or end on a charge the field lines can only enter the surface and end inside it if there is a charge inside the surface for them to end on. This means that regardless of what the arrangement of charges is like outside the surface it is only the charge inside the surface that determines the total flux.

So when you are using Gauss's law you need not worry about anything outside the surface. Gauss's law tells us that the total flux is proportional only to the charge inside the surface.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.