A small conducting spherical shell with inner radius a and outer radius b is con

Question

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 of -1q and the outer shell has a total charge of +3q.

Select True or False for the following statements.
The total charge on the outer surface of the large shell is +2q.
The electric field in the region r > d is zero.
The total charge on the inner surface of the small shell is -4q.
The electric field in the region c < r < 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 -1q.
The electric field in the region r < a is zero.

Use Gauss's Law

Explanation / Answer

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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 theinner 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 + 2q because the sphere itself must have a charge of +2q

now we have the inner sphere complete

now since the inner sphere has a charge of +2q

we can put a surface anywhere between the two surfaces and get a flux related to +2q 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 +2q charges from the first sphere in order to have 0 flux we must have 0 charges total inside the gausian sphere we made..

Qnet = +2q + Qinnersurface

since Qnet = 0

0 = +2q + Qinnersurface
Qinnersurface = -2q

so this shows that the inner surface of the outer sphere must be a charge of -2q

so we have a charge of +2q on the inner spheres outter surface and -2q 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 = +2q and +4q were given that means the total net charge inside the system is +6q

this means we need a net charge on the outersphere of +4q but we have a charge of -2q on the inner surface of it ...

mathematically

Qnet = +4q
Qinner = -2q
Qouter = ?

Qnet = Qinner + Qouter
+4q = -2q + Qouter

Qouter = +4q +2q
= +6q

so the total charge on the outersurface of the outer sphere is +6q

that creates the necessary +4q charge on the outer sphere
that makes the necessary charge of +6 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 = +2q outer side
Qc = -2q innerside
Qd = +6q outer side

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