Home » Ask & Discuss » Physics. » Electricity « Back to Discussion

Electricity

Blazing goIITian

Joined:	12 Apr 2008
Post:	424

4 Oct 2009 19:33:33 IST

0 People liked this

357

FIND ELECRIC FIELD......

None

Share this article on:

Comments (5)

zzzzz

Blazing goIITian

Joined: 12 Apr 2008

Posts: 424

4 Oct 2009 21:35:55 IST

Like 0 people liked this

a thin glass rod is bent into a semicircle of radius R. a charge +q is uniformly distributed along d upper half nd a charge -q is uniformly distributed along d lower half as shown in d above figure. find d electric field at d centre of semi-circle

PANKAJ KUMAR

Cool goIITian

Joined: 6 May 2009

Posts: 66

4 Oct 2009 21:57:22 IST

Like 0 people liked this

by symmetry we can say dat d horizontally rightward component of upper half nd horizontally leftwad component of lower half cancel each other...nd net electric field wil be vertically downward......we cn find it by integratin...do u hv problem in integration...

AViK

Blazing goIITian

Joined: 5 Sep 2008

Posts: 953

6 Oct 2009 19:10:25 IST

Like 0 people liked this

Am getting.... E = Q / e₀(pi*R)²

Priyak Dey

Scorching goIITian

Joined: 24 Sep 2009

Posts: 266

6 Oct 2009 21:57:57 IST

Like 1 people liked this

By symetry we see that all the horizontal components get cancelled by the electric

field due to the counterpart on the other ring...

CONSIDERING 1 PART OF THE RING....

Let linear charge density=L=Q/[pi*R/2]=2Q/[pi*R].

Now dECos@ = 1/[4*pi*e₀]* L*dl/R²...dl=elementary length=Rd@...@=theta

Nw integrate to get total E from a limit of '0' to 'pi/2'..

replace the value of L=2Q/[pi*R].

Now TOTAL ELECTRIC FIELD DUE TO THE WHOLE RING = 2*E...

Therefore,,

E_TOTAL=1/[4*pi*e₀]* Q/[pi*R].....

WE ARE MULTIPLYING BY @ AS WE TAKE IN CONIDERATION BOTH THE PARTS OF THE RING...MAGNITUDE OF E WILL BE SAME DUE TO BOTH..-ve SIGNB ONLY HELPS IN DETERMINING DIRECTION OF E....

CHEERZ.....RATE IF U FIND DIS USEFUL...........

Sagar Saxena

Forum Expert

Joined: 8 Oct 2008

Posts: 7221

13 Oct 2009 14:23:32 IST

Like 1 people liked this

hello dear

from the above figure it is clear that the the components of the electric field along the x axis will get cancelled

x axis component::

+X axis by by +Q side of the ring

-X axis by -Q side of the ring

hence will get cancel with each other

for vertical field component:

-Y axis by the +Q side of the ring

-Y axis by the -Q side of the ring

hence it will get add up::

hence by indicdual 1/4th ring the electric field is::

E = K (lamda) / R

hence net is along Y axis: it is 2E

where lamda = Q / (pi.R/2)

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team