# VEKTORANALYS Kursvecka 6 övningar. PROBLEM 1 SOLUTION A dipole is formed by two point sources with charge +c and -c Calculate the flux of the dipole.

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VEKTORANALYS Kursvecka 6 övningar

PROBLEM 1 SOLUTION A dipole is formed by two point sources with charge +c and -c Calculate the flux of the dipole field on a closed surface S that (a)encloses both poles (b)encloses only the plus pole (c)does not enclose any pole - + Vector field from a point source located in r 0 : Vector field from two point sources located in r 1 and r 2 If the origin is outside V If the origin is inside V Flux from a point source: (a) 4c4c-4  c =0 (b) 4c4c 0 = 4  c (c) 00 = 0 (a) (b) (c) 1

S V PROBLEM 2 SOLUTION Use the Gauss theorem to calculate the flux of the vector field: on the surface S: x y z The field is singular at  =0 (the z-axis) The Gauss theorem cannot be applied on S!!! We divide V into 2 volumes: V=V 0 +V  Thin cylinder with radius  along the z-axis The boundary surface to V 0 is: Does not contain the z-axis  here we can apply Gauss! VV SS SS SS 2

=0 Integrals III and IV are zero: I II III IV Integrals II is: On S ,  3

But our cylinder is very “thin”, which means that we are interested in Integrals I is: The volume is a sphere with centre in the point (0,0,2) Therefore, to use spherical coord. we must do a coordinate transformation: z’=z-2 0 4

LAPLACE OPERATOR IN CURVILINEAR COORDINATES The Laplace equation is The Laplace operator is the divergence of the gradient: CARTESIAN COORDINATES GRADIENT DIVERGENCE CYLINDRICAL COORDINATES GRADIENT DIVERGENCE SPHERICAL COORDINATES GRADIENT DIVERGENCE CARTESIAN CYLINDRICAL SPHERICAL 5

PROBLEM 3 SOLUTION A cylinder with radius R has a charge density  c. Calculate the potential and the electric field: (a)Inside the cylinder (b)Outside the cylinder Assume that the cylinder is infinitely long and that the potential on the surface is V 0. Due to the symmetry of the problem, the solution will depend on the radius only:  Inside the cylinder =0 Because the solution depends only on  cylindrical coord. The equation becomes: 6

=0 Because the solution depends only on  Divergent at  =0 NOT physical!  a=0 Outside the cylinder And the equation becomes: Multiplying by  Integrating in  Dividing by  Integrating in  Multiplying by  Integrating in  Dividing by  Integrating in  There is no charge 7

Now we must determine the three integration constants b, c and d. We have three conditions: (1)Continuity of the electric field at  =R (2) The potential at  =R (3) Continuity of the potential at  =R (1) (2) (3) =0 8

  R rr  R V0V0 9

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