A community for students.
Here's the question you clicked on:
 0 viewing
choupi
 2 years ago
Is voltaege zero for a conductor in electrostatic eq? I was thinking no, but since V=Ed it seems like there could be a difference in V by virtue of the particles being a certain distance apart. Could anyone explain Thanks!
choupi
 2 years ago
Is voltaege zero for a conductor in electrostatic eq? I was thinking no, but since V=Ed it seems like there could be a difference in V by virtue of the particles being a certain distance apart. Could anyone explain Thanks!

This Question is Open

ghazi
 2 years ago
Best ResponseYou've already chosen the best response.0@choupi it's change in voltage = E.change in distance that is \[dv=E.dx or E= dv/dx\] so what you are asking is bit unclear...but as far as i can understand it's just drifting of electrons under the influence of electric field that gives change in potential

r0oland
 2 years ago
Best ResponseYou've already chosen the best response.0The electric field inside a conductor is always zero (for electrostatics), see: http://ocw.mit.edu/courses/physics/802scphysicsiielectricityandmagnetismfall2010/conductorsandinsulatorsconductorsasshields/MIT8_02SC_notes9.pdf "Module 9: Conductors and Insulators" on page 16. Hence, the potential difference in that conductor is always zero. The whole conductor can however have a potential difference in respect to a reference point. This depends on the reference (eg. infinity where E=0) as well as the charge on and the geometry of the conductor.

Kumar777
 2 years ago
Best ResponseYou've already chosen the best response.0good question E=dv/dx since the electricfield inside the coductor is always zero you can make out that the potential inside a conductor is always constant.IT NEVER CHANGES.1 conductor 1 potential.Therefore potential difference inside a conductor is always zero.

harshaash
 2 years ago
Best ResponseYou've already chosen the best response.0take a gaussian surface,just inside the conductor.now gauss law says \[\int\limits_{?}^{?}E.dA=(Const)q _{enclosed}\]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...is this the answer to your ques?well there could arise a question from this reasoning...ie..we proved that.\[\int\limits_{?}^{?}E.dA\] is 0 but it may not necessarily imply E is 0....but that can also be proved as E is a consv field.....

harshaash
 2 years ago
Best ResponseYou've already chosen the best response.0it was not a contradiction..in fact your questing was not clear do u mean outside the conductor or inside? outside bearing no ext charges,E also 0 for the same reason..gauss law(when i meant,in the above reasoning that,take a gaussian surfacejust inside the conductor it means just inside its surface) but if there were ext charges E MAY not be 0

harshaash
 2 years ago
Best ResponseYou've already chosen the best response.0well a practical application of above is..... u r advised to sit in cars on a rainy day,as any amount of thunder,may not effect you as you are in a closed metallic shell
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.