# NOTES3 - 9/18/02 Examples of Gausss Law Example Determine...

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9/18/02 Examples of Gauss’s Law Example Determine the net electric flux out of the each of the surfaces shown in the figure. Solution: 1 S dQ ⋅= DS v 2 0 S Q = v 3 S Q Q Q + = v 1 S Q Q Q Q 2 S 3 S

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It is interesting to note that the integral form of Gauss’s law can be used to solve for E or D in problems where there is a high degree of symmetry (i.e., the field does not vary along at least 2 coordinate directions). The next three examples demonstrate the application of Gauss’s law to find E associated with three different charge distributions. Example Find the electric field due to a uniformly distributed sphere of charge ( 0 VV ρ = is a constant for r a ). Solution: Use Gauss’s law , V SV dd V ⋅= ∫∫ DS v . By symmetry ˆ r D = Dr , where D r is only a function of r . Thus, assuming r a y z x a a r S
22 0 00 000 2 3 2 0 2 2 2 2 ˆ sin sin 4 sin ( ) 3 sin 4 4 a V rr rV r r r rd d r d r d d a Dr d d Q Q DrQ Q D r ππ θ θφ ρ π θθφ ρ θθφ = ⋅= == = = = ∫∫ ∫∫∫ or 3 0 2 [V/m] 3 V r o a E r ε = . Note that this is the same D field as that for a point charge 3 0 4 3 V Qa = at the origin. For r a , 0 3 2 0 0 ˆ sin sin 4 4 3 3 r V V r V r d r d r d d r r D ρπ = = = or 3 [V/m] 4 r o Qr E a πε = . a r S

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Example Find the electric field due to an infinite uniform line charge 0 ρ = AA . Solution: Use Gauss's law, S dd ⋅= ∫∫ DS A A A v . By symmetry ˆ D = D ρ , where D is only a function of . Thus, 1 ˆ S dS ρρ = ⋅+ D ρ D 2 ˆ () zh S dS =− ⋅− + zD 3 ˆ S dS d =+ z A A A 2 0 0 0 0 (2 )2 2 [V/m] 2 hh o Dd z d d z Dh h E π ρφ πρ πε ρ −− = = = A A A This solution agrees with our previous result. x y 2 h z 2 S 1 S 3 S 0 A
Static Charges and Conductors Within and on conductors, charges can move with little resistance to motion. If free charges are placed on or within a conductor so as to cause a net charge imbalance, there will be an electric field established between these like charges. This electric field will exert a force on the individual charges causing them to move. The charges will tend to repel one another, moving as far apart as possible and arranging themselves such that they no longer experience any force. Thus they eventually reach a static condition in which there is no net force on any charge and therefore no electric field: = E 0

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## This note was uploaded on 08/05/2011 for the course ECE 2317 taught by Professor Staff during the Spring '08 term at University of Houston.

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NOTES3 - 9/18/02 Examples of Gausss Law Example Determine...

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