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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 May 2008 20:58:08 IST
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can the feild due to a conducting cone having a charge Q be found using gauss law? how?
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 May 2008 22:22:37 IST
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step no 1. draw a gaussian surface like the given surface.
step 2. get its area
step 3. apply theorem to get E
N0W I DON'T KNOW THE AREA OF A CONE.
There must be some formula.
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altitude begets altitude. |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 May 2008 23:30:56 IST
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@paulparthapratim HEY DUDE. CAN I ASK U WHICH TYPE OF GAUSSIAN SURFACE U R USING.AS FAR AS I KNOW THE ELECTRIC FIELD ON EACH POINT ON ANY GAUSSIAN SURFACE MUST BE SAME AND HENCE IT MUST BE EQUIDISTANT FROM THE CONE BUT I REALLY CANT THINK OF ANY SUCH TYPE OF SURFACE.SO, CAN U PLZ.. TELL ME WHICH TYPE OF GAUSSIAN SURFACE U R USING/
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Right point indrajeet. While gauss law is valid for such a charge distribution, it is not useful because we cannot convert integral of E.ds into E times some kind of area. We can't find field due to cone by gauss's law because of absence of symetry.
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Krishna Gopal Singh
B.Tech Chemical Engg
IIT Delhi 2002
Currently doing PhD from IIT Delhi |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 14 May 2008 10:09:35 IST
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yes sir u r right.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 14 May 2008 10:33:38 IST
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indrajee_bariar,u r wrong.
it is not necessary that electric field at every point on a gaussian surface should be same.the only required condition for a gaussian surface is that the surface should be closed.
if the electric field at every point on a gaussian surface is same,then why do we take closed integral of E.dS.
if E is same ,then E should come out of integral sign in the basic gauss's theorem equation itself.( {INT(E.dS)}=q/e0 should become {E.[INT(dS)]=q/e0}
it only comes out of integral sign in special cases like cylinder,sphere etc where field is uniform.
*e0=epsilon zero.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 14 May 2008 10:45:33 IST
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now we know that electric field at a point just outside the conductor for is E=(e0)/(surface charge density)
if u have resnick and halliday refer the section of a charged isolated conductor in the chapter gauss law.
the derivation is given.or u can refer any good book and see abt electric field outside a charged isolated conductor in gauss law chapter.
this is valid only for a point just outside a conductor.for faraway points from the conductor,it ceases to hold true.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 14 May 2008 13:12:32 IST
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@praveenkumar OK THEN TELL ME HOW WILL U CALCULATE THE ELECTRIC FIELD IF U CANT TAKE E OUT OF THE INTEGRAL ASSUMING IT TO BE CONSTANT
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 14 May 2008 13:56:58 IST
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sabne resnick padha hoga
suno bhai......gauss law is gven in every era gera buk.......sabhi buk main....... and resnick haliday is a gud buk also....... ....... bhai if u remember u shud hve read ki integration of E.dS is not easy to carry in areas where either E is not constant or some specialty abt it is not gven....... ab yeh poore saaf saaf likha hai wahan pe........ yeh toh yaad hoga......if nahin toh phir se padhle bhai........ E shud be const. over the gaussian surface..... or any face of it ....
aur ek baat ......... to calculate field at infinte distance....... then ats the use of takin it as a cone....... aise hi sperical point particle ho gaya
hope i m clear
in the cone E is not constant...... so dun apply it...... forget abt S .... thats after thng which is also not convinient to deal wid........
Cheers!!!!!
Cheers!!!!!
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