- #1

tellmesomething

- 324

- 38

- Homework Statement
- None

- Relevant Equations
- None

My teacher said that gauss law may not accounts for the field due to the outside charges in the LHS in this expression

##\int E.ds## = ##\frac{q}{\epsilon}## as field lines coming in the surface leave it as well. Hence the total flux is 0

However i dont think thats very consistent with the official definition of gauss law which states that the field due to both the inside and outside charges should be taken into account.

But his argument makes sense. Please explain why Its taken into account (if it is) and why not (if not)

One fallacy I found in his argument was the fact that equal number of field lines enter and leave the closed surface.. Isnt that wrong? If ive got an isolated point charge outside isnt the field changing with distance therefore at a distance x from the charge when its entering The closed surface its strength is different compared to a distance x+y where its leaving the closed surface..

Is my reasoning wrong?

##\int E.ds## = ##\frac{q}{\epsilon}## as field lines coming in the surface leave it as well. Hence the total flux is 0

However i dont think thats very consistent with the official definition of gauss law which states that the field due to both the inside and outside charges should be taken into account.

But his argument makes sense. Please explain why Its taken into account (if it is) and why not (if not)

One fallacy I found in his argument was the fact that equal number of field lines enter and leave the closed surface.. Isnt that wrong? If ive got an isolated point charge outside isnt the field changing with distance therefore at a distance x from the charge when its entering The closed surface its strength is different compared to a distance x+y where its leaving the closed surface..

Is my reasoning wrong?