How electric field lines going out from surface is zero?

1. Sep 12, 2015

gracy

in the video above the instructor explains from time 8:22 to 8:55 that how and why electric field lines going out from the surface which do not enclose charge is zero but I don't understand then why not electric field lines going out from the spherical surface ( enclosing charge) is zero.Because by applying what the instructor said in this case also ,electric field lines going out comes out to be zero.

2. Sep 12, 2015

blue_leaf77

Here is the correct way to determine whether an electric field line at a particular point $\mathbf{r}$ is considered to be "going out" or "going in":
- if $\mathbf{E}(\mathbf{r}) \cdot d\mathbf{a} > 0$, the electric field line at this point is leaving the closed surface
- if $\mathbf{E}(\mathbf{r}) \cdot d\mathbf{a} < 0$, the electric field line at this point is going inside the closed surface
- if $\mathbf{E}(\mathbf{r}) \cdot d\mathbf{a} = 0$, the electric field line at this point grazes the surface, it doesn't count in the flux traversing the surface
The surface element $d\mathbf{a}$ is by convention taken to be directed outward from the closed surface.
In your picture, it's obvious that the quantity $\mathbf{E}(\mathbf{r}) \cdot d\mathbf{a}$ is positive at all points in the spherical surface.

3. Sep 13, 2015

gracy

But applying what you said

but it should be leaving the closed surface
so
but in my case it is

4. Sep 13, 2015

blue_leaf77

At that exact point where $\mathbf{E}(\mathbf{r}) \cdot d\mathbf{a} = 0$, the line does not leave nor enter the surface. If you keep tracing this line you may find it enter or leave the surface at another point.
Are you referring to your last picture, with $d\mathbf{a}$ denoted by that area vector to the right? That vector looks almost parallel with the electric field line at the same point, doesn't it? E(r)⋅da=0 cannot be zero then.

Last edited: Sep 13, 2015
5. Sep 13, 2015

gracy

I Understand now.Thanks a lot.Really really helpful.