2. E -field of a hollow sphere Question: Calculate E -field in arbitrary points outside the sphere Available: A hollow sphere, radius R , with surface charge density [C/m 2 ]
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4. Analysis and Symmetry 2. Coordinate axes: Z-axis = polar axis 3. Symmetry: spherical 4. Spherical coordinates: r, 1. Charge distribution: (surface charge) C/m 2 ] Z X Y R R R e r e e
5. Analysis, field build-up 4. E i,xy , E i,z 5. expect: E i,xy = 0, to be checked !! 6. E = E z e z only ! 1. XYZ-axes X Y R Z 2. Point P on Z-axis . P 3. all Q i ’ s at r i , i , i contribute E i to E in P r Q i E i e r
6. Approach to solution dQ = dA r and ( e r .e z ): see next page Z e r r X Y . P e z d d dQ at dA Distributed charge: dQ dE dA=(R.d R. sin d R. sin
7. Calculations (1) dA=(R.d R. sin d r and ( e r .e z) : see next page Z X Y e r . P r dQ at dA dE d d e z R. sin
8. Calculations (2) r 2 = ( R. sin + (z P - R.cos ( e r .e z ) = (z P - R.cos r Z X Y e r . P r dQ at dA dE d d e z R. sin r R e r e z z P R.sin R.sin R.cos z P - R.cos
9. Calculations (3) dA=(R.d R. sin d r 2 = ( R. sin + (z P - R.cos ( e r .e z ) = (z P - R.cos r Z X Y e r . P r dQ dE d d e z R. sin
10. Calculations (4) result for E in P: z P < R : E = 0 Z X Y e r . P r dQ dE d d e z R. sin z P > R :
11. Conclusions for homogeneous charge distribution: total charge seems to be in center the end r < R : E = 0 E =0 r > R : E E