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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.@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.I think inner surface pe induce hoga
outer wle pr kyu nhi bro?
Andar -q or bahar +q ?
The flux is zero inside the spherical shell so the net charge is zero and hence -q charges appears in the inner surface thru induction....
*outside, the flux is zero outside
Why outside ?
Flux inside shellA is zero as there is no charge in it, on surface positive as given, now shellB is earthed so it's potential is zero, now write potentials on surface of shellB due to both the shells:
KQa/Rb + KQb/Rb = 0 (Rb is radius of shellB, and Qa is the +ve charge given on shellA) so Qa = -Qb, ehich means same magnitude of -ve charge induced on shellB, so will it be inside or outside? Let's see... Now take the inner radius of shellB (Rb') and outer radius (Rb'') now find potential on the oter surface, which is zero due to Earthing:
KQa/Rb'' + KQb/Rb'' + KQ''/Rb'' = 0 (Q'' on outer surface of shellB)
So KQa/Rb" + KQb/Rb" cancel out as Qa = -Qb... So we're left with KQ"/Rb" = 0, K and Rb" can't be zero, so Q" is zero which proves charge on outer shell is zero
So -ve charge must be induced on the inner surface
If earthing wasn't there there would be equal magnitude+ve charge on the outer surface of shellB aswell
Bahar nhi hoga +q
Read what i said above
Got it bro
Thanks a lot
close please
+solved @Aguilar
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