A nonconducting sphere 1.0 m in diameter with its center on the x axis at x = 4

Question

A nonconducting sphere 1.0 m in diameter with its center on the x axis at x = 4 m carries a uniform volume charge of density = 3.3 µC/m3. Surrounding the sphere is a spherical shell with a diameter of 2.0 m and a uniform surface charge density  = -1 µC/m2. Calculate the magnitude and direction of the electric field at the following locations.

(a) x = 4.4 m, y = 0

_____I N/C ______J N/C

(b) x = 4 m, y = 0.9 m

______ I N/C _______ J N/C

(c) x = 2.0 m, y = 2.0 m
_______ I N/C _______ J N/C



I really don't know where to begin this problem, if you can please help me get started, i would highly appreciate it. Thanks.

Explanation / Answer

diameter of nonconducting sphere d1 = 1 m radius of the sphere R1 = 0.5 m diameter of the spherical shell d2 = 2 m radius of the spherical shell R2 = 1 m volume charge density =  3.3 µC/m3 surface charge density = -1 µC/m2 electric field due to nonconducting shere and spherical shell is        E = E1 + E2     .................. (1) where , E1 = electric field due to nonconducting shere              E2 = electric field due to spherical shell a) at x = 4.4 m and y = 0 m (inside the spherical shell) : the electric field, inside the spherical shell is zero. so , at this point E2 = 0 the electric field E1 = (4/3)(kd) i^      where , d = 4.4 m - 4 m                   = 0.4 m the electric field E1 = (4/3)(3.14*9*109*3.3*10-6)(0.4) i^                              = (49.73*103 N/C) i^ substitute these values in eq (1) , we get        E = (49.73*103 N/C) i^ + 0           =  (49.73*103 N/C) i^ + 0 j^
direction :                 = tan-1 (0/49.73*103)                    = 00 ...................................................................... b) ...................................................................... at x = 4 m and y = 0.9 m (inside the spherical shell) : the electric field, inside the spherical shell is zero. so , at this point E2 = 0 for x = 4 m : the electric field E1 = (4/3)(kd') i^      where , d' = 4 m - 4 m                   = 0 m the electric field E1 = (4/3)(3.14*9*109*3.3*10-6)(0) i^                              = 0 for y = 0.9 m : electric field E2 = (4/3y2)[k(R1)3] j^                         = (4*3.14/3(0.9)2)[9*109*3.3*10-6*(0.5)3] j^                         = (57.567*103 N/C) j^ substitute these values in eq (1) , we get        electric field E = 0 i^ + (57.567*103 N/C) j^         direction :                 = tan-1 (57.567*103 /0)                   = 900
.............................................................................
........................................................................... at x =2 m and y = 2 m , charge qshell = Ashell                     = [4R22]                     = (-1*10-6)(4)(3.14)(1)2                     = -12.56*10-6 C charge of the sphere is                qsphere = Vsphere                           = [(4/3)R13]                           = (3.3*10-6)(4/3)(3.14)(0.5)3                           = 1.727*10-6 C    if x = 2 and y = 2 ,     r = 22 + 22 = 8 = 2.828 m unit vector r^ = -2/2.828 i^ + 2.828/2 j^                      = -0.7071 i^ + 0.7071 j^ ..............................................................................      electric field E shell is      E2 = E shell = [k(qshell)(r^)] / r2                        = [9*109*(-12.56)*10-6*(-0.7071 i^ + 0.7071 j^)] / (2.828)2                                    = [9.99*103 i^ - 9.99*103 j^] N/C      electric field Esphere is   E1 = Esphere = [k(qsphere)(r^)] / r2                      = [9*109*1.727*10-6*(-0.7071 i^ + 0.7071 j^)] / (2.828)2                                  = [-1.3742*103 i^ + 1.3742*103 j^]  N/C substitute these values in eq (1) , we get         E = 8.615*103 i^ - 8.615*103 j^
direction :         = tan-1 (-8.615*103/8.615*103)            = -450 ............................................................................... answers : a) E =  (49.73*103 N/C) i^ + 0 j^ b) E = 0 i^ + (57.567*103 N/C) j^     c) E = 8.615*103 i^ - 8.615*103 j^                                   the electric field, inside the spherical shell is zero. so , at this point E2 = 0 for x = 4 m : the electric field E1 = (4/3)(kd') i^      where , d' = 4 m - 4 m                   = 0 m the electric field E1 = (4/3)(3.14*9*109*3.3*10-6)(0) i^                              = 0 for y = 0.9 m : for y = 0.9 m : electric field E2 = (4/3y2)[k(R1)3] j^                         = (4*3.14/3(0.9)2)[9*109*3.3*10-6*(0.5)3] j^                         = (57.567*103 N/C) j^ substitute these values in eq (1) , we get        electric field E = 0 i^ + (57.567*103 N/C) j^         direction :                 = tan-1 (57.567*103 /0)                   = 900
.............................................................................
........................................................................... at x =2 m and y = 2 m , charge qshell = Ashell                     = [4R22]                     = (-1*10-6)(4)(3.14)(1)2                     = -12.56*10-6 C charge of the sphere is                qsphere = Vsphere                           = [(4/3)R13]                           = (3.3*10-6)(4/3)(3.14)(0.5)3                           = 1.727*10-6 C    if x = 2 and y = 2 ,     r = 22 + 22 = 8 = 2.828 m unit vector r^ = -2/2.828 i^ + 2.828/2 j^                      = -0.7071 i^ + 0.7071 j^ ..............................................................................                           = 1.727*10-6 C    if x = 2 and y = 2 ,     r = 22 + 22 = 8 = 2.828 m unit vector r^ = -2/2.828 i^ + 2.828/2 j^                      = -0.7071 i^ + 0.7071 j^ ..............................................................................      electric field E shell is      E2 = E shell = [k(qshell)(r^)] / r2                        = [9*109*(-12.56)*10-6*(-0.7071 i^ + 0.7071 j^)] / (2.828)2                                    = [9*109*(-12.56)*10-6*(-0.7071 i^ + 0.7071 j^)] / (2.828)2                                    = [9.99*103 i^ - 9.99*103 j^] N/C      electric field Esphere is   E1 = Esphere = [k(qsphere)(r^)] / r2                      = [9*109*1.727*10-6*(-0.7071 i^ + 0.7071 j^)] / (2.828)2                                  = [-1.3742*103 i^ + 1.3742*103 j^]  N/C substitute these values in eq (1) , we get         E = 8.615*103 i^ - 8.615*103 j^
direction :         = tan-1 (-8.615*103/8.615*103)            = -450 ............................................................................... answers : a) E =  (49.73*103 N/C) i^ + 0 j^ b) E = 0 i^ + (57.567*103 N/C) j^     c) E = 8.615*103 i^ - 8.615*103 j^                                                        = [9*109*1.727*10-6*(-0.7071 i^ + 0.7071 j^)] / (2.828)2                                  = [-1.3742*103 i^ + 1.3742*103 j^]  N/C substitute these values in eq (1) , we get         E = 8.615*103 i^ - 8.615*103 j^
direction :         = tan-1 (-8.615*103/8.615*103)            = -450 ............................................................................... answers : a) E =  (49.73*103 N/C) i^ + 0 j^ b) E = 0 i^ + (57.567*103 N/C) j^     c) E = 8.615*103 i^ - 8.615*103 j^                                   ............................................................................... answers : a) E =  (49.73*103 N/C) i^ + 0 j^ b) E = 0 i^ + (57.567*103 N/C) j^     c) E = 8.615*103 i^ - 8.615*103 j^         b) E = 0 i^ + (57.567*103 N/C) j^     c) E = 8.615*103 i^ - 8.615*103 j^                             

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