Consider the following arrangement of two conducting hollow spheres with a point
ID: 2193833 • Letter: C
Question
Consider the following arrangement of two conducting hollow spheres with a point charge of Q0 = 3.90 ?C at the center. The inner sphere has a radius of 0.011 m and carries a net charge of Q1= -2.50 ?C. The outer sphere has a radius of 0.061m and carries a net charge of Q2 = 6.90 ?C.a) Calculate the magnitude of the electric field at point A located at a distance 0.021 m from the center.
b) Calculate the surface charge density on the outer surface of the outer sphere. Neglect the thickness of the sphere.
Explanation / Answer
(a) In this part we have our observation point between the two hollow spheres. As the shells are conducting and situation described is radially symmetric, we can be sure that charges (whatever they may be) on the inner and outer surface of sphere uniformly are distributed i.e. we can use gauss law to find the electric field. Or one can recall the results for conducting shells which states that the inside the conducting shell electric field is zero (unless there a net charge, obiviously) and outside it behaves as if there is a charge of same magnitude sitting at the center of shell. Using this result we can see electric field at 21mm (0.021m) would just be electric field due to charge at center + due the first shell (Note that outside shell will not contribute as it charge distribution is symmetric).
E(r = 21 mm) = kQ/r2 = 6 x 105 N/C
where k =9x109 N-m2 / C2
Q = Q0 + Q1 = 1.4 C
(b) Now to get surface charge density we need the total charge before we can divide it with the surface area. Inside a conductor electric field is zero. This is a constrain that we apply in order to know the surface charge on the outer surface of the outer sphere. Consider the first shell. At inside surface of the shell we see that -Q0 charge consequently Q0 apeares on the outer surface of the first shell. Total charge at the outer surface of first shell is Q0 + Q1. By applying the same reasoning for outer shell, we clearly see that the Total charge at the outer surface of outer surface is Q0 + Q1 + Q2 = 8.3 C.
Hence the charge density = Qtot/4r2 = 2.23 x 10-3 C/m2
where Qtot = 8.3 C and r = 61 mm
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