Electric flux related to perpendicular suf?

[hermtm2]

Electric flux related to perpendicular suf?

Hello.

This is my first in here.
I am studying on Physics electricity and magnetism.
While I am reading the concept of electric flux, the textbook repeatly mention about "perpendicular to the surface" . Why is the perpendicular surface important?

Here is the one of the paragraph from the book.

"For a sphere centered on the point charge Q, the flux is just this product pi=EA=Q/e0, as introduced in the expression for Coulmb's law above. This expressition pi is correct in this case because the electric field at the surface is always perpendicular to the surface and its magnitude is the same at all points on the surface."

Thanks.

 

[Bob S]

If your sphere is a conductor, then there is no electric field parallel to the surface.

 

[Born2bwire]

That's really just the definition of flux, the integral over a surface of the vector quantity that is normal to the surface. It is a measure of flow through a surface area. If the flow is parallel to a surface, then there is no flow through the surface, illustrated by the fact that the normal component is zero.

 

[Thevanquished]

electric field lines is ALWAYS perpendicular to the surface it hits and also does not cross one another. Thats just a fact

 

[espen180]

electric field lines is ALWAYS perpendicular to the surface it hits and also does not cross one another. Thats just a fact

You are right in that field lines (or surfaces) never intersect, at least in conservative fields, as that would violate the conservation of energy.

Still, electric field lines need not be perpendicular with every surface. I think you are talking about $$\int \vec{E}\cdot \vec{dl}=0$$ at the surface of any conductor. Indeed the electric field lines are perpendicular to all conducting surfaces, which simply means that conductor surfaces are equipotential surfaces.

 

[dx]

The component of the flow which is parallel to the surface does not contribute to the flux (because it is along the surface), so the flux depends only on the perpendicular component of the flow.

 

[Thevanquished]

You are right in that field lines (or surfaces) never intersect, at least in conservative fields, as that would violate the conservation of energy.

Still, electric field lines need not be perpendicular with every surface. I think you are talking about $$\int \vec{E}\cdot \vec{dl}=0$$ at the surface of any conductor. Indeed the electric field lines are perpendicular to all conducting surfaces, which simply means that conductor surfaces are equipotential surfaces.

haha yea forgot that it was only for conductors cause questions usually deal with conductors. thanks for the correction though, point noted

 

[RohansK]

The electric or magnetic field essentially associates a force experienced by a point charge or a magnetic substance placed in the field. This strength is called the flux.

Rather the force experienced is the outcome or result or what we can call a manifestation of the exsistence of any such field.

And any force experienced is the maximum only when it acts Normal or Perpendicular to a surface. And the force exerted by the field can be defined only for the maximum intensity of it for a given charge producing the field.

And as mentioned earlier this can happen only when the force is Perpendicular ie inturn the flux being Normal to the surface on which it acts.

Thus the field is always perpendicular to the surface.

 

[dx]

Nonsense, RohansK. The field is in general not perpendicular to a gaussian surface. It is only true in special situations, like the one in the OP where the surface is a sphere centered on a charge.

 

[Doc Al]

While I am reading the concept of electric flux, the textbook repeatly mention about "perpendicular to the surface" . Why is the perpendicular surface important?

In special situations of high symmetry (like the one referenced in that quoted paragraph) the electric field is perpendicular to the Gaussian surface and has the same magnitude everywhere on the surface. That makes it easy to compute the total flux through that surface in terms of the electric field and thus use Gauss's law to solve for the electric field.

 

[mikelepore]

In special situations of high symmetry (like the one referenced in that quoted paragraph) the electric field is perpendicular to the Gaussian surface and has the same magnitude everywhere on the surface. That makes it easy to compute the total flux through that surface in terms of the electric field and thus use Gauss's law to solve for the electric field.

When you say "high symmetry" you must mean a point charge or a spherical distribution. If you put a Gaussian surface around a line distribution or sheet distrubution it will be necessary for the closed surface to have some segments where the electric field is parallel and not perpendicular.

 

[Doc Al]

When you say "high symmetry" you must mean a point charge or a spherical distribution. If you put a Gaussian surface around a line distribution or sheet distrubution it will be necessary for the closed surface to have some segments where the electric field is parallel and not perpendicular.

I didn't want to go off into those details in answering the OP's question, but you are of course correct. You need enough symmetry to make the flux easy to calculate. That will happen when the field is perpendicular or parallel to the surface. (It's real easy to calculate the flux contribution when the field is parallel to the surface--it's zero.)

 

[espen180]

This is the reason why carefully choosing your gaussian suface is important. Choosing the wrong surface, where high symmetry does not occur, might render the task of calculating the flux near impossible (or at least extremely laborious).