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Class_12new

# Class_12new - PHYS809 Class 12 Notes Potential of grounded...

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1 PHYS809Class12Notes Potential of grounded spherical conductor in a uniform external electric field The potential for a uniform field E 0 is 0 0 . φ = - E r Since there is azimuthal symmetry about the axis through the center of sphere that is parallel to the direction of the external field, we look for an external solution of form ( ) ( ) ( ) 0 0 1 1 1 0 0 cos cos cos . l l l l l l l l B B P P E rP r r φ θ φ θ θ + + = = = + = - Since the potential is zero on the sphere, which we take to have radius a , we see that B l = 0 except for B 1 . Hence ( ) 3 3 0 0 1 0 2 2 cos cos . E a a E r P r E r r φ θ θ = - = - This is the same as the solution found using the method of images. Solution using expansion on the symmetry axis An important result for an azimuthally symmetric solution of Laplace’s equation is that it completely determined from the potential on the symmetry axis, which we will take to be the z -axis. On the symmetry axis ( ) ( ) ( ) ( ) 1 0 1 0 , 0, 1 , 0.

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Class_12new - PHYS809 Class 12 Notes Potential of grounded...

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