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Gauss law a n I i t can be thought of as aching the u be of electric field lines going through a surface which depends upon the angle between the surface and the electric field A KE A nose Gauss law can be formulated as ThefIof through any Feds surface is equal to the net charge geneleo enclosed by this surface follows dit quale and 1 Note 2 Note 3 Note 1 Gauss I Gaussian surface law holds for any surface one particular surface the proble of deferring the that is closed Usually makes electric field easy when calculating the net change inside a closed surface we take accord of the sign of the charge when applying Gauss closed surface charges outside the surface law for a ignore the wa Caulu Dstavus Gaussla Both laws describe the relationship between electric charge ed the electric field derive Couto b s law fro Gauss hav One ed versa can vice free that the electric e is exerted this force zero velocity E lase a are assure that the a aero vanishing force The electric field inside a conductor We shall prove that electric field inside conductor vanishes Expericels have shown that in general contain lo ductors charges electrons negative lat s to move field is not In such a non by the field on ead electron would on salt that moving electrons We will have the following properties I both have not been observed in conductors they heal the conductor 2 they generate magnetic fields in a see non L et d 1791 1867 MichaelFarady 1791 1867 such as Experiment effect cannot penetrate show that the Faraday cage electric field also a conductor o o o 0 add An isolated changed conductor with a cavity a cavity to a solid conductor charges redide on the walls We can The question can of the cavity We If surface which is conductor surface the electric field is O to using Gauss law to get answer we we apply Gauss law to a Gaussian located just below the Inside the conductor are Thus lo GE AE From Gauss law u have 9 we Il the love compare the upper equation with we get gene O Conclusion There is no charge all outer surface of the conductor onthe cavity walls the excess charge q re airs on the A changed isolated conductor was can reside conilade that no charges conductor charged Ne can inside a Assure a conductor that Any charges inside the conductor that have save sign repell each other That will the force that will tend to separate lead to a them as mad as possible If electrons are free to move they will concentrate on the surface to maximise their distance Here no additional charges there are du I inside the conductor ed Gauss law to conclude that the electric tied inside the conductor is 0 can we s use Recipe for applying Gauss law 1 Make a sketch of the charge distribution Identify the symmetry of the charge 2 distribution and electric field Gauss law is true for any closed that makes surface S the calculation of the flax integral as its effect on Choose one the easy as possible B 4 Us Gauss electric field vector law to determine the j 7 So dit I surface of asphery A HITv2 0 9 d E A E E E ut hiya I Divertion of E Point charge q qg the symmetry of the Gaussian surface Fro the electric field has to point radially outwards ut f i Hollowsphet changed hollow sphere 10 I de I dit LA 4Th E Ultra Qf at E E direction as above E points radially outwards uts E a Hollow sphere Gaussian surface inside the hollow sphere So d D 17 4 To YE E 4Th E E O Electric field generated by a log uniforly dad rod n II 9 iii f dit de Since the enclosing angle the electric field is O they do not contribute to the integral A 21T h E A E 2Ish of sea d L E ZITA K G Eu E rites Direction of i The direction of can be concluded fro the superposition principle It points outwards the rod Es s Electric field generated by a thin infinite inducting unify charged sheet 0 Fro am syne try the electric to thy sheetand pointaway fo it Un choose a cylindrical Gaussian surface S with the caps of area S on either side ofthe sheet We divide S into three sections I S is the cap onthe rightside Sa is the carved side Sz is the cap on the leftside of to to flax of E Aline The nut 0 3 0 42 since the electric field is parallel to the surface 2 E Acivile Gauss law tells us of 9 7 T f 2 E Age Taft E Direction of Using superposition we see that the electric field always points perpendicular away fro the plane

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SUNY Geneseo EDUC 488 - Gauss Law

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