Physics 2102 Gabriela Gonz lez Physics 2102 Gauss law Carl Friedrich Gauss 1777 1855 Electric Flux r r E dA Electric Flux A surface integral CLOSED surfaces define the vector dA as pointing OUTWARDS Inward E gives negative F Outward E gives positive F Gauss Law Consider any ARBITRARY CLOSED surface S NOTE this does NOT have to be a real physical object The TOTAL ELECTRIC FLUX through S is proportional to the TOTAL CHARGE ENCLOSED The results of a complicated integral is a very simple formula it avoids long calculations S Surface q E d A 0 One of Maxwell s 4 equations Properties of conductors Inside a conductor in electrostatic equilibrium the electric field is ZERO Why Because if the field is not zero then charges inside the conductor would feel forces and would be moving the system is not in equilibrium yet SO the electric field pushes charges in a conductor to redistribute themselves until they make the field inside the conductor ZERO Excess charges are always on the surface of the conductors but never inside Example Faraday s Cage Given a hollow conductor of arbitrary shape Suppose an excess charge Q is placed on this conductor Suppose the conductor is placed in an external electric field How does the charge distribute itself on outer and inner surfaces a Inner Q 2 outer Q 2 q b Inner 0 outer Q E d A 0 Surface c Inner Q outer 0 Choose any arbitrary surface inside the metal Gauss law since E 0 flux 0 Hence total charge enclosed 0 All charge goes on outer surface http www mos org sln toe Inside cavity is shielded from all external electric fields Faraday Cage effect Gauss Law Example A spherical conducting shell has an excess charge of 10 C A point charge of 15 C is located at center of the sphere Use Gauss Law to calculate the charge on inner and outer surface of sphere R2 a Inner 15 C outer 0 b Inner 0 outer 10 C c Inner 15 C outer 5 C Does the answer change if the 15C is not at the center R1 15 C More Properties of conductors We know the field inside the conductor is zero and the excess charges are all on the surface The charges on the surface produce an non zero electric field outside the conductor On the surface of conductors in electrostatic equilibrium the electric field is always perpendicular to the surface Why Because if not charges on the surface of the conductors would move with the electric field Gauss Law Example Charged conductor of arbitrary shape no symmetry non uniform charge density What is the electric field near the surface where the local charge density is s Applying Gauss law we have A AE 0 E 0 Solving for the electric field we get E 0 THIS IS A GENERAL RESULT FOR CONDUCTORS Gauss Law Example Spherical symmetry q E d A 0 Consider a POINT charge q pretend that you don t know Coulomb s Law Surface Use Gauss Law to compute the electric field at a distance r from the charge r Use symmetry q draw a spherical surface of radius R E centered around the charge q E has same magnitude anywhere on surface E normal to surface E A E 4 r 2 q 0 F q e0 q kq E 2 2 2 A 4p r 4pe0 r r Electric fields with spherical symmetry shell theorem The field outside a spherical distribution of charge is the same as if all the charge is at the center of the sphere A conducting spherical shell has a charge of 10C and a point charge of 15C at the center What is the magnitude of the electric field produced E E k 15C r2 If the shell is conducting 10 C 15C E 0 E k 5C r2 r What is the electric field if the shell is a uniformly charged sphere Gauss Law Planar symmetry Infinite plane with uniform charge density s E is NORMAL to plane Construct Gaussian box as shown Applying Gauss law q A we have 2 AE 0 0 Solving for the electric field we get E 2 0 Two infinite planes E s 2e0 E s 2e0 Q Q E s e0 A uniform field E 0 E 0 Insulating and conducting planes Q E 2 0 2 A 0 Q Insulating plate charge distributed homogeneously Q 2 Q E 0 2 A 0 Conducting plate charge distributed on the outer surfaces Gauss Law Cylindrical symmetry Charge of 10 C is uniformly spread over a line of length L 1 m Use Gauss Law to compute magnitude of E at a perpendicular distance of 1 mm from the center of the line Approximate as infinitely long line E radiates outwards Choose cylindrical surface of radius R length L co axial with line of charge E 1m R 1 mm Gauss Law cylindrical symmetry cont Approximate as infinitely long line E radiates outwards Choose cylindrical surface of radius R length L co axial with line of charge E 1m E A E 2 RL q L 0 0 L E 2k 2 0 RL 2 0 R R R 1 mm Compare with exact calculation L 2 x k L k r 2 2 2 r x r L 2 r r 2 L 2 2 L 2 dx E y k r 2 2 3 2 r x L 2 l Q L r if the line is infinitely long L r k L 2k Ey 2 r r L 2 Summary Gauss law provides a very direct way to compute the electric flux In situations with symmetry knowing the flux allows to compute the fields reasonably easily Spherical field of a spherical uniform charge kqins r2 Uniform field of an insulating plate s 2e0 of a conducting plate s e0 Cylindrical field of a long wire 2kl r Properties of conductors field inside is zero excess charges are always on the surface field on the surface is perpendicular and E s e0
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