http://capa-new.colorado.edu/msuphysicslib/Graphics/Gtype53/prob11_condshells.gi

Question

http://capa-new.colorado.edu/msuphysicslib/Graphics/Gtype53/prob11_condshells.gif

Regarding to this picture: http://capa-new.colorado.edu/msuphysicslib/Graphics/Gtype53/prob11_condshells.gif A small CONDUCTING spherical shell with inner radius a and outer radius b is concentric with a larger conducting spherical shell with inner radius c and outer radius d. The inner shell has a total charge -1q and the outer shell has a total charge +4q. Which of the following statements are true: (Give ALL correct answers, i.e., B, AC, BCD...) The total charge on the inner surface of the small shell is zero. The electric field in the region r less than a is zero. The total charge on the outer surface of the large shell is +3q. The electric field in the region c less than r less than d is zero. The electric field in the region r greater than d is zero. The total charge on the inner surface of the large shell is zero. The total charge on the outer surface of the small shell is -5q.

Explanation / Answer

these are usually difficult to do but they can usually be done with gauss's law..

so if we put a spherical surface inside the inner surface of the sphere we will get 0 flux so that means there is 0 CHARGE INSIDE THE FIRST sphere which is obvious..

if we put a spherical surface between the inner surface and the outer surface of the inner sphere we must have a flux of 0 because its a conducting surface... and these are static charges... so again the inner surface must have a charge of 0

since the inner surface of the inner sphere has a charge of 0 that means the outer surface of the inner sphere MUST have a charge of - 1q because the sphere itself must have a charge of -1q

now we have the inner sphere complete

now since the inner sphere has a charge of -1q

we can put a surface anywhere between the two surfaces and get a flux related to -1q this is because the inner sphere will be inside any surface that is generated between the inner sphere and the outer sphere..

we can put a surface between the inner surface of the outer sphere and the outer surface of the sphere and the flux must be 0 BECAUSE its inside the conducting sphere with static charges..

this means that since we KNOW we have -1q charges from the first sphere in order to have 0 flux we must have 0 charges total inside the gausian sphere we made..

Qnet = -1q + Qinnersurface

since Qnet = 0

0 = -1q + Qinnersurface
Qinnersurface = +1q

so this shows that the inner surface of the outer sphere must be a charge of +1q

so we have a charge of -1q on the inner spheres outter surface and +1q on the outer spheres inner surface..
this adds up to 0 total and this makes 0 flux inside the outer sphere

===============

final part we put a gaussian sphere outside the outer sphere..

since the net charges = -1q and +4q were given that means the total net charge inside the system is +3q

this means we need a net charge on the outersphere of +4q but we have a charge of +1q on the inner surface of it ...

mathematically

Qnet = +4q
Qinner = +1q
Qouter = ?

Qnet = Qinner + Qouter
+4q = +1q + Qouter

Qouter = +4q -1q
= +3q

so the total charge on the outersurface of the outer sphere is +3q

that creates the necessary +4q charge on the outer sphere
that makes the necessary charge of +3 q for the total system of the two spheres.

and that is the semi easy but can be complicated way to find out what the charges are ..

to summarize

Qa = 0 inner side
Qb = -1q outer side
Qc = +1q innerside
Qd = +3q outer side

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