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sonofa_nh
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A charge of +3q is placed at the center of an unchanged conducting shell. What will be the charges on the inner and outer surfaces of the shell, respectively?
A. 3q, +3q
B. 3q, +6q
C. 3q, 3q
D. q, +q
 2 years ago
 2 years ago
sonofa_nh Group Title
A charge of +3q is placed at the center of an unchanged conducting shell. What will be the charges on the inner and outer surfaces of the shell, respectively? A. 3q, +3q B. 3q, +6q C. 3q, 3q D. q, +q
 2 years ago
 2 years ago

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hosein Group TitleBest ResponseYou've already chosen the best response.2
i think a is correct
 2 years ago

sonofa_nh Group TitleBest ResponseYou've already chosen the best response.0
how sure are u?
 2 years ago

hosein Group TitleBest ResponseYou've already chosen the best response.2
from conservation charge principle
 2 years ago

sonofa_nh Group TitleBest ResponseYou've already chosen the best response.0
excellent...ty
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.0
More intuitively perhaps, a charge brought near a conductor induces in the conductor a charge of opposite sign closest to the charge. So the inside of the sphere should have an induced charge which is negative, because +3q is positive. Now, the exterior must be the opposite of whatever is in the interior. Hence its sign is +.
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.0
Therefore option A or D is correct. Now the only question is whether you think that induced charge will have the same magnitude or not?
 2 years ago

hosein Group TitleBest ResponseYou've already chosen the best response.2
same hence shell is uncharged
 2 years ago

prakharJ Group TitleBest ResponseYou've already chosen the best response.0
just use the gauss theorem here. Take a closed spherical Gaussian surface with lie within the thickness of conducting shell. Since the electric field inside the conductor is zero here, so according to gauss law which is \[closed \int\limits E.dS = q(enclosed)/\epsilon\] . Since E is zero q enclosed must also be zero, so to do that charges(3q here) from conducting shell comes to the inner side to make the net charge inside that gaussian surface zero.. And also due to conservation of charge , on the outer side +3q develops.
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.0
Exactly right.
 2 years ago
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