In Figure 17.25, a conducting sphere ofradius r1 = 0.050 m is placed at the cent

Question

In Figure 17.25, a conducting sphere ofradius r1 = 0.050 m is placed at the center of a sphericalconducting shell of inner radius r2 = 0.100 m and outer radiusr3 = 0.140m. The inner sphere carries a charge of -4.0 ×10-9 C. The outer spherical shell carries a net charge of 3.0× 10-9 C. Calculate the magnitude of the electric field atthe following distances from the center of the spheres:

(a)   r = 0.075 m (in the air spacebetween spheres),

(b)   r = 0.120 m (in the metal ofthe spherical shell), and

(c)   r = 0.200 m (outside thespherical shell).

In Figure 17.25, a conducting sphere ofradius r1 = 0.050 m is placed at the center of a sphericalconducting shell of inner radius r2 = 0.100 m and outer radiusr3 = 0.140m. The inner sphere carries a charge of -4.0 10-9 C. The outer spherical shell carries a net charge of 3.0 10-9 C. Calculate the magnitude of the electric field atthe following distances from the center of the spheres: (a) r = 0.075 m (in the air spacebetween spheres), (b) r = 0.120 m (in the metal ofthe spherical shell), and (c) r = 0.200 m (outside thespherical shell).

Explanation / Answer

In Figure 17.25, a conducting sphere ofradius r1 = 0.050 m is placed at thecenter of aspherical conducting shell of inner radius r2 = 0.100 m andouter radius r3 = 0.140m. The innersphere carries a charge of -4.0 × 10-9 C. The outer spherical shell carries a netcharge of 3.0 × 10-9 C.   Calculatethe magnitude of the electric field at thefollowing distances from the center ofthespheres: Applying Gauss's Law to a spherical surface of radius "R" and concentric with the 2 given spheres, we obtain:      E•dS = (Q_enclosed)/     ---->   E*4*Pi*R^2 = (Q_enclosed)/     ---->   E = (Q_enclosed)/(*4*Pi*R^2)     ---->   E = (8.9876e+9)*(Q_enclosed)/R^2 (a)   r = 0.075 m (in the air space betweenspheres),      R = 0.075 m      (Q_enclosed) = -4.0e-9 C       E = (8.9876e+9)*(Q_enclosed)/R^2           = (8.9876e+9)*(-4.0e-9)/(0.075)^2           = -6391 V/m      {E Magnitude} = 6391 V/m (b)   r = 0.120 m (IN themetal of the spherical shell).       R = 0.120 m      Electric field "E" is always zero (0) insideconductor.       E = 0 V/m      {E Magnitude} = 0 V/m (c)   r = 0.200 m (outside the sphericalshell).       R =  0.200m      (Q_enclosed) = (-4.0e-9 C) + (3.0e-9C) = -1.0e-9C       E = (8.9876e+9)*(Q_enclosed)/R^2           = (8.9876e+9)*(-1.0e-9)/(0.200)^2           = -224.7 V/m      {E Magnitude} = 224.7 V/m .

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