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@Gyro Gearloose
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.ik by logic that all charge on sphere will travel to outermost part of shell to make field inside 0 and potential uniform
but how do i prove it mathematically
like q' charge travels from sphere to shell
i mean u can equate the potential at point A (on the solid sphere) to that at point B the end of the wire
lets assume q' to be the charge remaining on the solid sphere. -q' is induced on the inner side and q on the outer side cuz total charge has to remain q
then.. at A = kq'/a - kq'/2a + kq/3a.. at B -kq'2a + kq/3a + kq'/2a
so q'=0
which means all the charge goes to the outer surface
isnt this what u wanted to prove @hardcoreisdead
so u are sort of considering one thick shell as combination two thin spherical shells
shouldnt Vb = kq"/2a+kq/3a + kq/3a
yeah cuz it's a think conducting shell. there's no field in the bulk of the conductor
what's q''?
charge
ik that.. i meant where
like i took q' on the solid sphere
mb ill rewrite
shouldnt Vb = kq/3a+kq'/3a - kq'/3a
why tho
i dont see it..
for any charge inside a shell , potential due to it on the surface of shell = k*charge/radius of sphere right
q' and -q' are both inside the outermost shell
and q resides on outermost shell
so in denominator only 3a should be there
that gives q'=q'/2 which basically means q'=0
same result but the right way (ig?)
Net field inside the conducting shell at any point must be zero. Only config that does that is if all the charge is on the surface. That is the mathematical proof.
i may be quite off but ig this is the potential at a point on the outer surface of the shell, right? the one i wrote was for a point on the inner surface of the shell (directly to which the wire is connected)
we will equate the potentials of the points joined by the wire.. dont know if i understand u properly
i was looking for eqns and stuff
OHH RIGHT
Gauss' Law
V= kq'/2a -kq'/2a + kq/3a
right that makes sense
Field inside the conductor is identically zero, so flux must be zero for any spherically symmetric Gaussian surface
yeah ofc thats usefull but i just wanted a lil concept check
In fact for any Gaussian surface
Which doesn't cross the outermost sphere
yeah
+solved @ns
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