For the configuration shown in the figure below, suppose a = 5.00 cm, b = 20.0 c

Question

For the configuration shown in the figure below, suppose a = 5.00 cm, b = 20.0 cm, and c = 25.0 cm. Furthermore, suppose the electric field at a point 17.0 cm from the center is measured to be 3.65 times 10^3 N/C radially inward and the electric field at a point 50.0 cm from the center is of magnitude 172 N/C and points radially outward. From this information, find the following. (Include the sign of the charge in your answer.) the charge on the insulating sphere the net charge on the hollow conducting sphere the charge on the inner surface of the hollow conducting sphere the charge on the outer surface of the hollow conducting sphere

Explanation / Answer

Given

   insulating sphere of radius a = 5 cm, hallow conducting sphere inner radius b = 20 cm, outer raius c = 25 cm

and electric field measured at r = 17cm is (between a,b) is E1 = 3.65*1063 N/C which is radially inward, and

       at r2 = 50 cm outside the conducting sphere is E2 = 172 N/C radially out ward

a) charge on the insulating sphere is

   E1 = kQinsulator/r1^2

   ==> Q_insulator = E1*r1^2/k

           = 3.65*10^3*0.17/9*10^9 C

           = - 6.894*10^-8 C (radially inward)

b)

charge on the hallow conducting sphere
      

       E2 = kQtotal/r2^2

       Q total = E2*r2^2/k

           = 172*0.5^2/(9*10^9) C

       = 4.78*10^-9 C

   charge on the conductor = Q_total - Q_insulator

               = 4.78*10^-9 + 6.894*10^-8 C

               = 7.372*10^-8 C

c) charge on the inner surface of the hallow sphere is

equal to the charge on the insulator = 6.894*10^-8 C

d)

outer surface of the hallow sphere is = Q_conductor - Q inner conductor

               = 7.372*10^-8 - 68.94*10^-9 C
  

                   = 4.78*10^-9 C

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