How electric field lines going out from surface is zero?

AI Thread Summary
Electric field lines from a surface that does not enclose charge are zero, while those from a spherical surface that does enclose charge are not. The determination of whether an electric field line is leaving or entering a surface relies on the dot product of the electric field and the area vector. If this product is greater than zero, the field is leaving the surface; if it is less than zero, it is entering. At points where the product equals zero, the field line neither enters nor exits the surface. Clarification of these concepts helps in understanding electric field behavior around charged surfaces.
gracy
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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.

el.png
 
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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.
 
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But applying what you said
area vector.png

blue_leaf77 said:
if E(r)⋅da=0\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
but it should be leaving the closed surface
blue_leaf77 said:
if E(r)⋅da>0\mathbf{E}(\mathbf{r}) \cdot d\mathbf{a} > 0, the electric field line at this point is leaving the closed surface
so
blue_leaf77 said:
if E(r)⋅da>0
but in my case it is
blue_leaf77 said:
E(r)⋅da=0
 
gracy said:
but it should be leaving the closed surface
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.
gracy said:
but in my case it is
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.
 
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I Understand now.Thanks a lot.Really really helpful.
 
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