Phys.219_Lec._04._2012.

Phys.219_Lec._04._2012. - Lecture 4-1 Physics 219 Question...

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Lecture 4-1 Where (if any) is the net electric field due to the following two charges equal to zero? Q Q a Physics 219 – Question 1 – January 23, 2012. a) at (-a,0) b) at (2a,0) c) at (a/2,0) d) at (0,a) and (0,-a) e) nowhere x y
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Lecture 4-2 Where (if any) is the net electric potential due to the following two charges equal to zero? Q Q a Physics 219 – Question 2 – January 23, 2012. a) at (-a,0) b) at (2a,0) c) at (a/2,0) d) at (0,a) and (0,-a) e) nowhere x y
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Lecture 4-3 Gauss’s Law: Qualitative Statement Form any closed surface around charges Count the number of electric field lines coming through the surface, those outward as positive and inward as negative . Then the net number of lines is proportional to the net charges enclosed in the surface .
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Lecture 4-4 Electric flux General definition of electric flux: Ei i i E A  (must specify sense , i.e., which way) To state Gauss’s Law in a quantitative form, we first need to define Electric Flux. # of field lines N = density of field lines x area” Area is a vector. is the magnitude of the area. The direction of is perpendicular to the surface . The flux is : = cos E i i i i ii AA EA    Projected = A
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Lecture 4-5 Why are we interested in electric flux? E is closely related to the charge(s) which cause it. If we now turn to our previous discussion and use the analogy to the number of field lines, then the flux should be the same even when the surface is deformed. Thus should only depend on Q enclosed.  2 2 0 4 E kq E AR R q   Consider Point charge q A
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Lecture 4-6 Gauss’s Law: Quantitative Statement 0 enclosed E Q  Must use a closed surface. Take the unit normal vector outward always.
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Lecture 4-7 Flux of an Electric Field The flux through a surface of arbitrary shape in an electric field can be computed by summing up fluxes through infinitesimal segments of the surface EA   • segment 1 makes negative contribution.
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Phys.219_Lec._04._2012. - Lecture 4-1 Physics 219 Question...

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