Step 1 We observe that there is spherical symmetry in this charge distribution. This mea cal symmetry. At any point at which the magnitude of the electric field is non-zero, the magnitude will depend only on the distance r from the center of the distribution, and the electric field vector will be in the radial direction. To take advantage of this symmetry, we choose a erical gaussian surface of radius / centered on the center of the charge distribution as shown in the diagram. For such a gaussian surface, the magnitude of the electric field is the same ough the Gauss n electric field of constant magnitude E is given by PE = EA = 4RP-E. Gauss's law states that the total electric flux through any closed surface is Qinsade/for where the constant zo = 8.85 x 10-12 C?/N - m? is the permittivity of free space and Oinside is the net charge enclosed by the Gaussian surface. Therefore, we have *= = 417 = JInside and the magnitude of the electric field at all points on the Gaussian surface is 2inside - KeQingid E = - where the Coulomb constant k. = = = 8.99 x 109 N - m?/C?. (a) When r = 1.00 cm, all points on the Gaussian surface are located within the con aterial of the solid sphere of radius 2.00 cm. a nonzero electric field cannot exist within a con we conclude that E = [ o N/C at all points where r = 1.00 cm. Step 2 (b) For r = 3.00 cm, all points on the Gaussian surface are located in the space between the solid sphere and the inner surface of the spherical shell. The net charge enclosed by this Gaussian surface is that on the in = +6.61 AC, so the ctric field at this location is E = keginside 8.99 x 109 N . m7/C3) X Your response differs from the correct answer by more than 10%. Double check your calculations. x 10- C) v x10-2 m) X Your response differs from the correct answer by more than 10%. Double check your calculations. x 10' N/C. Since Qinside > 0, at r = 3.00 cm, E =[ Your response differs from the correct answer by more than 109%. Double check your calculations, x 107 N/C radially [outward . Submit|Skip (you cannot come back)A solid conducting sphere of radius 2.00 cm has a charge of 9.30 AC. A conducting spherical shell of inner radius 4.00 cm and outer radius 5.00 cm is concentric with the solid sphere and has a charge of -3.24 uC. Find the electric field at the following radii from the center of this charge configuration. (a) r = 1.00 cm magnitude N/C direction --Select- (b) r = 3.00 cm magnitude N/C direction -Select- c) r = 4.50 cm magnitude N/C direction --Select- r = 7.00 cm magnitude N/C direction -Select