Electric Field Zero on Gaussian Surface with No Charges

[tecnics]

why is the electric field zero , if the closed gaussian surface doesn't enclose any charges.
but,, if the charges are present at a distance outside the conductor??

and why isn't the electric field zero inside the gaussian surface, if there are charges distributed on the surface of the gaussian surface..

assume that our gaussian surface is an enclosed sphere...

 

[Matterwave]

Your questions seem very confused, you mention a conductor...what conductor?

I'm going to venture a guess that you're confused by Gauss's law. Gauss's law says that the surface integral of the electric field over a closed surface is equal to the charge enclosed by this closed surface.

If there is no charge enclosed by the Gaussian surface (which doesn't have to be a physical surface like a conductor, but just any closed surface you want to draw) then the surface integral of the electric field is 0. It doesn't say that the electric field is 0, it says that the integral of the electric field over the Gaussian surface is 0. In the language of field lines, if there are no enclosed charges, then every field line that enters the Gaussian surface must exit it as well, so that the total integral is 0.

 

[tecnics]

Your questions seem very confused, you mention a conductor...what conductor?

I'm going to venture a guess that you're confused by Gauss's law. Gauss's law says that the surface integral of the electric field over a closed surface is equal to the charge enclosed by this closed surface.

If there is no charge enclosed by the Gaussian surface (which doesn't have to be a physical surface like a conductor, but just any closed surface you want to draw) then the surface integral of the electric field is 0. It doesn't say that the electric field is 0, it says that the integral of the electric field over the Gaussian surface is 0. In the language of field lines, if there are no enclosed charges, then every field line that enters the Gaussian surface must exit it as well, so that the total integral is 0.

ok////
can u pls explain the Gauss law? i am definitely confused by it..

 

[Matterwave]

Sure.

Gauss's law states, quite simply, that the surface integral over a closed surface of the electric field in space (let's just treat the vacuum for now, we can modify it if you're in a medium) equals the enclosed charge inside the surface. In mathematical language the law is (using SI units):

$$\oint_S \vec{E}(x,y,z) \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}$$

Let's break this expression down. First you start off knowing that there is some electric field that permeates space (the electric field might turn out to be 0, if there are no charges anywhere). It is a vector field (it associates a vector with each point in space, even if that vector might be 0).

Then you pick ANY closed surface (the surface must be closed, so you can't pick a plane or something, but you can pick a sphere, or a box, or a cylinder) and you take the integral of the electric field over this surface. When you are taking this integral, this is a surface integral, so you have to first take the dot product of the electric field vector with the surface element vector (which is a vector that is normal to the surface itself). This dot product tells you by how much the electric field "pierces" your surface at that point. You add up all these contributions from all over your enclosed surface, and you will obtain a single number (with units of V*m). Gauss's law says that this number is the same as the number you get if you take the charge enclosed by the surface you drew and divide it by the permittivity of free space (an artifact of the SI system of units, but let's not worry about that right now).

Now, Gauss's law is completely general. You can pick any shaped surface you want (as long as it's closed) and any charge distribution you want and Gauss's law will still hold. However, unless your problem has some good symmetries involved (e.g. spherical symmetry, or cylindrical symmetry), it is very difficult, in practice, to actually perform that surface integral! You want to pick your Gaussian surfaces to match the symmetry of the given charge distribution. So if the charge is a point charge, choose a spherical Gaussian surface. If the charge is an infinite plane, then choose a pillbox.

 

[pmacias]

The electric field is defined as the force per unit charge experienced by a test charge. In the case of a closed Gaussian surface with no charges enclosed, the net force on any test charge placed inside the surface would be zero. This is because the electric field is a result of the presence of charges, and without any charges within the surface, there is no electric field.

If there are charges present at a distance outside the conductor, these charges do not contribute to the electric field inside the Gaussian surface. This is because the electric field lines originating from these external charges would pass through the surface without intersecting it, therefore not affecting the electric field inside.

On the other hand, if there are charges distributed on the surface of the Gaussian surface, the electric field would not be zero inside the surface. This is because the charges on the surface would create an electric field that would intersect the surface and contribute to the net electric field inside. This is similar to how a charged conductor would have an electric field inside, even though the net charge of the conductor is zero.

In the case of an enclosed sphere as our Gaussian surface, the electric field inside would be zero if there are no charges inside the sphere or if the charges inside are uniformly distributed. This is because in both cases, the electric field lines would be symmetrically distributed and cancel each other out. However, if the charges inside are not uniformly distributed, the electric field inside the sphere would not be zero as there would be a net electric field resulting from the asymmetry of the charge distribution.

Overall, the electric field inside a closed Gaussian surface is dependent on the charges enclosed within the surface and their distribution. In the absence of enclosed charges, the electric field would be zero.