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I-5

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I-5 Special Electrostatic Fields 11. 7. 2003 1 Main Topics • • • • Electric Charge and Field in Conductors. The Field of the Electric Dipole. Behavior of E. D. in External Electric Field. Examples of Some Important Fields. 11. 7. 2003 2 A Charged Solid Conductor I • Conductors contain free charge carriers of one or both polarities. Charging them means to introduce in them some excess charges of one polarity. • A special case are metals : • every atom which joins metal structure, often crystallic, keeps some of its electrons in its vicinity but the valence electrons, which are bounded by the weakest forces, are shared by the whole structure and they are the free charge carriers. They can move within the crystal when electric (or other) force is acting on them. • It is relatively easy to add some excess free electrons to metal and also to take some out of it. 11. 7. 2003 3 A Charged Solid Conductor II • Adding electrons means charging the metal negatively. • Taking some electrons out means charging it positively. • For our purposes we can consider the ‘holes’ left after missing electrons as positive free charge carriers each with charge +1e. • So effectively the charged metal contains excess charges either negative or positive, which are free to move. 11. 7. 2003 4 A Charged Solid Conductor III • Excess charges repel themselves and since they are free to move as far as to the surface, in equilibrium, they must end on a surface. • In equilibrium there must be no forces acting on the charges, so the electric field inside is zero and also the whole solid conductor must be an equipotential region. 11. 7. 2003 5 A Hollow Conductive Shell I • In equilibrium again: • the charges must remain on the outer surface. • the field inside is zero and the whole body is an equipotential region. • The above means the validity of the Gauss’ law. • To proof that let’s return to the Gauss’ law. 11. 7. 2003 6 The Gauss’ Law Revisited I • Let us have a positive point charge Q and a spherical Gaussian surface of radius r centered on it. Let us suppose radial field: kQ E (r )  p r • The field lines are everywhere parallel to the outer normals, so the total flux is:  e  E (r ) A   0 Qr 1 2 p • But if p2 the flux would depend on r ! 11. 7. 2003 7 The Gauss’ Law Revisited II • The validity of the Gauss’ law  p = 2. • By using a concept of the solid angle it can be shown that the same is valid if the charge Q is anywhere within the volume surrounded by the spherical surface. • By using the same concept it can be shown that the same is actually valid for any closed surface. • It is roughly because from any point within some volume we see any closed surface confining it under the solid angle of 4. 11. 7. 2003 8 A Hollow Conductive Shell II • Let first the shell be spherical. Then the charge density  on its surface is constant. • From symmetry, in the center the intensities from all the elementary surfaces that make the whole  surface always compensate themselves and E  0 • For any other point within the sphere they compensate themselves and E  0 only if p = 2. • Again, using the concept of solid angle, it can be shown, the same is valid for any closed surface. 11. 7. 2003 9 A Hollow Conductive Shell III • Conclusion: The existence of a zero electric field within a charged conductive shell is equivalent to the validity of the Gauss’ law. • This is the principle of: • experimental proof of the Gauss’ law with a very high precision: p – 2 = 2.7  3.1 10-16. • of shielding and grounding (Faraday’s cage). 11. 7. 2003 10 Electric Field Near Any Conducting Surface • Let us take a small cylinder and submerge it into the conductor so its axis is perpendicular to the surface. • The electric field • within the conductor is zero • outside is perpendicular to the surface • A non-zero flux is only through the outer cup   E 0 • Beware the edges!  is not generally constant! 11. 7. 2003 11 The Electric Dipole I • Materials can produce non-zero electric fields in their vicinity even when the total charge in them is compensated. • But they must contain so called electric multipoles in which the centers of gravity of positive and negative charges are not in the same point. • The fields produced are not centrosymmetric and decrease generally faster than the field of the single point charge. 11. 7. 2003 12 The Electric Dipole II • The simplest multipole is the electric dipole. • It is the combination of two charges of the same absolute value but different sign +Q and –Q.  • They are separated by vector l , starting in –Q. • We define the dipole moment as :   p  Ql 11. 7. 2003 13 The Electric Dipole II • Electric dipoles (multipoles) are important because they are responsible for all the electrical behavior of neutral matter. • The components of material (molecules, domains) can be polar or their dipole moment can be induced. • Interactions of dipoles are the basis of some types of (weaker) atomic bonds. 11. 7. 2003 14 Behavior of the Electric Dipole in External Electric Fields • In uniform electric fields the dipoles are subjected to a torque which is trying to turn their dipole moments in the direction of the field lines • In non-uniform electric fields the dipoles are also dragged. 11. 7. 2003 15 Some Examples • The field of homogeneously charged sphere • Parallel uniformly charged planes • Electrostatic xerox copier 11. 7. 2003 16 Homework • Now, you should be able to solve all the problems due Monday! 11. 7. 2003 17 Things to read • The first five lectures cover : Chapters 21, 22, 23 ! • Advance reading Chapter 24 - 1, 2, 3 • Who is really interested should try to see the physicist “Bible”: “The Feynman Lectures on Physics” 11. 7. 2003 18 The Solid Angle I • Let us have a spherical surface of radius r. From its center we see an element of the surface da under a solid angle d : da d  2 r We see the whole spherical surface under : 4r   2  4 r 2 The Solid Angle II If there is a point charge Q in the center the elementary flux through da is:   da cos  d e  E  da  E da cos   kQ 2 r Since the last fraction is d, the total flux is:  e  kQ d  kQ4  Q 0 ^ Intensities near more curved surfaces are stronger! • Let’s have a large and a small conductive spheres R, r connected by a long conductor and let’s charge them. Charge is distributed between them to Q, q so that the system is equipotential: 2 2 Q q a r Q r R R  ;  2     2 R r A R S R r r ^ Potential of Electric Dipole I • Let us have and a  a charge –Q at the origin  +Q in dl. What is the potential in r ? We use the superposition principle and the gradient:      ( r )    ( r )    ( r  dl )   kQ kQ kQ    (  dl )  grad ( ) r r r How to calculate grad(1/r)? • r is the distance from the origin : 1 2 2 2  12 r  x  y  z   (x  y  z ) r 2 2 2 • e.g. the first components of the gradient is : 2  12 3 ( x  y  y ) x  2 2 2 1 2  ( 2 )( x  y  y ) 2 x  3 x r 2 2 Potential of Electric Dipole II • The first two terms cancel:      kQdl  r kp  r  (r )   3 3 r r • The potential has axial symmetry with the dipole in the axis and axial anti-symmetry perpendicular to it. It decreases with 1/r2! ^ Electric Dipole - The Torque  • Let us have a uniform field with intensity E Forces on both charges contribute simultaneously to the torque: l T  2 QE sin  2 • The general relation is a cross product:    T  p E ^ Electric Dipole - The Drag • Let us have  a non-uniform field with intensity E and a dipole parallel to a field line (-Q in the origin). F  QE (0)  QE (dl )  dE  QE (0)  QE (0)  Qdl dx • Generally:    F  gradE  p ^ The vector or cross product I    Let c  a  b Definition (components) ci   ijk a j bk  The magnitude c    c  a b sin    Is the surface of a parallelepiped made by a , b. The vector or cross product II The vector c is perpendicular to the plane   made by the vectors a and b and they have to form a right-turning system.    ux uy uz  c  ax ay az bx by bz ijk = {1 (even permutation), -1 (odd), 0 (eq.)} ^