Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Can anyone try to help me, why is the electrical field 0 inside a sphere?

OCW Scholar - Physics II: Electricity and Magnetism
I got my questions answered at in under 10 minutes. Go to now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer


To see the expert answer you'll need to create a free account at Brainly

electrical field is not zero inside a sphere, it's zero inside conductors and inside the sphere that has charges on its surface! distribution on a spherical surface!
you can study Lecture 2- Electric Field and Dipoles, Walter L. for calculating and discussing about Field. and for conductors later lectures!
Thank you for the fast answer!

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

inside CONDUCTORS it is 0 for this reason take a gaussian surface,just inside the conductors gauss law says ∫E.dA=(Const)(Qenclosed) and Q enclosed is 0 we have E is all the conductors charges lie on the outer surface,as they would lie max dist apart.well there could arise a question from this proved that. ∫E.dA is 0 but it may not necessarily imply E is 0....but that can also be proved as E is a consv field.....
Thanks for your answers.
The charge placed at the centre or at a distance say r from the centre within a sphere has Area(E.da) = Q/e; e: epsilon. When u take a charge outside the sphere and if u calculate the directional fields entering and leaving the sphere it would be same. So the total electric field will sum upto zero. This is the point where Gauss law comes into the picture.
gauss's electric law,electric field line only works when intercepting a plane for only one time
You can easily understand the concept via energy analysis....... Imagine a sphere having uniform +ve surface charge density ,Now if you assume the sphere boundary to be thick enough to have materialistic volume ,then charges are of-course like charges ,therefore there will be repulsion between them and potential energy decreases. Now if the above sphere is non conducting(which doesn't allow any charge to pass through it's materialistic volume(which is negligible)) then charges present on inner side will keep residing on the same place and doesn't move even if our sphere's thick boundary is available(i.e. becoz it is now a insulator) Now if the above sphere is conducting(which allow any charge to pass through it's materialistic volume(which is negligible)) then charges present on inner side of boundary will move to outside surface of our conductor becoz of additional pathway provided by conduction body and potential energy decreases.Hence no excess charge will be on inner boundary of conducting body,net charge will be on outer surface of conducting body. |dw:1364799446002:dw| Apart form the explanation of Gauss law(∫E.dA=(Const)(Q{enclosed}) by using shell theorems from gravitation chapter u can also prove net electric field is zero inside conducting sphere.Becoz it is a spherical conductor for conductors \[\sigma r = const\] charge distribution will be symmetrical on surface of sphere and then u can apply shell theorems.

Not the answer you are looking for?

Search for more explanations.

Ask your own question