A small conducting spherical shell with inner radius a and outer radius b is con
ID: 2224426 • Letter: A
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 -2q and the outer shell has a total charge of +3q. http://loncapa2.physics.sc.edu/res/msu/physicslib/msuphysicslib/53_Efield_Calculus_Gauss/graphics/prob11_condshells.gif Select True or False for the following statements. The electric field in the region c < r < d is zero. The electric field in the region r > d is zero. The total charge on the inner surface of the large shell is -5q. The total charge on the outer surface of the large shell is +1q. The total charge on the inner surface of the small shell is -2q. The electric field in the region r < a is zero. The total charge on the outer surface of the small shell is zero.Explanation / Answer
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 + 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