4CONDUCTING SPHERICAL SHELL
E due to q' at O= 0
E exists due to q at P
So total E due to q at O = k q/4pir2
REASON : Charge will reside only on outer surface so as to take the electric field inside = 0 at every interior point.
4
UTTARA
·2010-03-31 03:30:36
E at P = kq'/R2 + kq/r2 ??? R --> Radius
(I'm not sure of this though)
1THE field inside IS NOT 0
since there is q' inside the sphere there will be -q' on the inner surface of the sphere.
the field inside will depend on q' and -q'(which is not uniformly distributed on the surface,charge distribution related to position of q' inside )
but not on q outside (electrostatic shielding ).
the outer surface of the sphere will have a net charge of q' which is again not distributed uniformly (charge distribution due to q outside , nothing to do with q' inside it ). The field at any point outside will depend on charge q at p and also on q' distributed on outer surface of sphere.
1
nish
·2010-04-02 11:44:22
in case 2 if u meant charge Q separately on both spheres , it will never happen.
the bigger sphere will hold twice as much charge as the smaller one .
then potential at midpoint will be potential of the combination (since the point lies inside the sphere,and potential is constant inside)
1See as far as the inner region is concerned , i totally agree with Nish .
One IMPORTANT THING TO NOTE IS : An outer charge CANNOT affect a charge kept inside a conducting sphere BUT an inner charge CAN INDIRECTLY affect the outer charge by inducing a charge on the outer surface of the spherical conductor and thereby it creates a field around itself [ which surely affects the outside charge also ].
1now..most important stuff here is the distribution of charges on the surface....
1)the external charge will first of all induce some charge on the surface such that the net field(that of the external charge + induced charge)=0
2)(due to gauss law)there will be an ununiform distribution of charges on the inner surface of the conductor....also(due to conservation of charge)due to this induced inner charge...the same magnitude of charge will appear on he external part of the conductor!!!!
Once all the charge distribution is clear all that is required to do is apply coulomb's law.....(just writing Coulomb's law for the charge as mentioned in 1) is diificult!!!! )
1
Grandmaster
·2010-04-03 01:08:32
though i am finding it a little difficult to draw the electric field lines.....
1
Grandmaster
·2010-04-03 01:10:35
here the midpoint is inside the larger shell......also we know that the field inside a conductor is 0.therefore the potential is that of the surface!!!
i.e. total potential at surface= pot. due to the charge on its surface + pot. due to the other shelll
= kQ/r + kQ/2r
1
Grandmaster
·2010-04-03 01:14:10
is there sumthing fishy about the third question......why will there be an elctric potential or field????
106arey nooo..
it should be grav. potential and field for third one
106
Asish Mahapatra
·2010-04-04 21:43:26
kaymant sir, please help here
I dont have answers so cant verify the correctness of what has been written above..
62see from http://www.cosmolearning.com/video-lectures/conductor-electrostatic-shielding/
38:00 to 42:40
4 minutes 40 seconds... will clear all the doubts we had about the 2nd part....
You have all the answer.. to the questions that we had yesterdya.. *(I was confused because i thiught that there is no effect.. the crrect thing is that there is no change in the effect due to change in postion of the charges etc..) but i guess you should see the video.. not see everything just the 4:40 mins that i have sent.. otherwise you will be spending too much time on other things...
