0
$\begingroup$

I have a sphere of radius "r" , the sphere has a total charge "q" distributed equally around the surface. A point "P" is at a distance "R" from the centre of the sphere.

I have a sphere of radius $r$ , the sphere has a total charge $q$ distributed equally around the surface. A point "P" is at a distance $R$ from the centre of the sphere. As per the book, to find the net flux from the gaussian sphere with radius $R$, we use ∅ = E × ds × cos (theta) , here , we are assuming the electric field at every particular point in the gaussian surface to be "E" as constant. And thus ,it can be easily calculated as net ∅ = E × 4πr²(the area vector makes 0° with E). But , in the picture , I can see the Electric field can be seen as the total electric field by the all the points on the surface made by a solid angle infront of the point , and only one field is normal to the area vector while the others make different angles. So why do we compute it with taking E as constant. And if I'm correct, how can I find the net flux of that particular point and ultimately for the whole gaussian surface?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Computing the total electric field of the sphere by adding up every contribution of every point is a very tedious process, precisely for the reason you mention.

Gauss law is a way to get the result much faster. First you prove that the field is radial ($\vec{E}=E(r)\,\vec{e}_\theta$) by studying the symmetries of the sphere, and then you can apply Gauss law to get the result as your book does.

When you apply Gauss, the strategy is in the choice of the "Gauss sphere" of radius $R$, precisely so that $\vec{E}$ and $d\vec{s}$ are parallel and $\theta=0$ everywhere.

Improve this answer
$\endgroup$
2
  • $\begingroup$ so, does this mean that whenever we find the net flux, of any region or the whole surface area , it is just an approximation tho we prove E to be radial , but practically it is not the only electric field point at a particular point on the surface? $\endgroup$
    – Aryan
    Commented Jun 7, 2022 at 2:08
  • $\begingroup$ In this case, it isn't an approximation, due to the unicity of the solution. There are some more complicated cases with less symmetries where approximations are necessary, like the field generated by a plane. If the plane is finite, the field isn't normal, so we usually say that, as long as we remain far from the edges, the plane is almost infinite so the field is almost normal. $\endgroup$
    – Miyase
    Commented Jun 7, 2022 at 6:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.