A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
Can anyone try to help me, why is the electrical field 0 inside a sphere?
anonymous
 4 years ago
Can anyone try to help me, why is the electrical field 0 inside a sphere?

This Question is Open

kmazraee
 4 years ago
Best ResponseYou've already chosen the best response.1electrical 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!

kmazraee
 4 years ago
Best ResponseYou've already chosen the best response.1you can study Lecture 2 Electric Field and Dipoles, Walter L. for calculating and discussing about Field. and for conductors later lectures!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thank you for the fast answer!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0inside CONDUCTORS it is 0 for this reason take a gaussian surface,just inside the conductors surface.now gauss law says ∫E.dA=(Const)(Qenclosed) and Q enclosed is 0 we have E is 0.now all the conductors charges lie on the outer surface,as they would lie max dist apart.well there could arise a question from this reasoning...ie..we 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.....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks for your answers.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0gauss's electric law,electric field line only works when intercepting a plane for only one time

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You 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 ofcourse 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.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.