ch4-Gauss’s Law | Electric Field | Quantity

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  Chapter 4 Gauss’s Law 4.1 Electric Flux............................................................................................................. 1 4.2 Gauss’s Law............................................................................................................. 2 Example 4.1: Example 4.2: Example 4.3: Example 4.4: Infinitely Long Rod of Uniform Charge Density ................................ 7 Infinite Plane of Charge.................................................................
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  Chapter 4Gauss’s Law 4.1 Electric Flux.............................................................................................................14.2 Gauss’s Law.............................................................................................................2Example 4.1: Infinitely Long Rod of Uniform Charge Density................................7Example 4.2: Infinite Plane of Charge.......................................................................9Example 4.3: Spherical Shell...................................................................................11Example 4.4: Non-Conducting Solid Sphere...........................................................134.3 Conductors.............................................................................................................14Example 4.5: Conductor with Charge Inside a Cavity............................................17Example 4.6: Electric Potential Due to a Spherical Shell........................................184.4 Force on a Conductor.............................................................................................214.5 Summary................................................................................................................234.6 Appendix: Tensions and Pressures........................................................................24 Animation 4.1: Charged Particle Moving in a Constant Electric Field..................25 Animation 4.2 : Charged Particle at Rest in a Time-Varying Field........................26 Animation 4.3 : Like and Unlike Charges Hanging from Pendulums.....................284.7 Problem-Solving Strategies...................................................................................294.8 Solved Problems....................................................................................................314.8.1 Two Parallel Infinite Non-Conducting Planes................................................314.8.2 Electric Flux Through a Square Surface.........................................................324.8.3 Gauss’s Law for Gravity.................................................................................344.8.4 Electric Potential of a Uniformly Charged Sphere.........................................344.9 Conceptual Questions............................................................................................364.10 Additional Problems............................................................................................364.10.1 Non-Conducting Solid Sphere with a Cavity................................................364.10.2 P-N Junction..................................................................................................364.10.3 Sphere with Non-Uniform Charge Distribution...........................................374.10.4 Thin Slab.......................................................................................................374.10.5 Electric Potential Energy of a Solid Sphere..................................................384.10.6 Calculating Electric Field from Electrical Potential.....................................380  Gauss’s Law 4.1   Electric Flux In Chapter 2 we showed that the strength of an electric field is proportional to the numberof field lines per area. The number of electric field lines that penetrates a given surface iscalled an “electric flux,” which we denote as  E  Φ . The electric field can therefore bethought of as the number of lines per unit area. Figure 4.1.1 Electric field lines passing through a surface of area  A .Consider the surface shown in Figure 4.1.1. Let ˆ  A = An  be defined as the area vector   having a magnitude of the area of the surface,  A , and pointing in the normal direction,. If the surface is placed in a uniform electric field E ˆ n  that points in the same directionas , i.e., perpendicular to the surface  A , the flux through the surface is ˆ n   ˆ  E  A EA Φ = ⋅ = ⋅ = EAEn   (4.1.1)On the other hand, if the electric field E  makes an angle θ  with (Figure 4.1.2), theelectric flux becomes ˆ n   n cos  E  EA E A θ  Φ = ⋅ = = EA  (4.1.2)where is the component of  E n ˆ  E  = ⋅ En   perpendicular to the surface. Figure 4.1.2 Electric field lines passing through a surface of area  A whose normal makesan angle θ  with the field.1  Note that with the definition for the normal vector , the electric flux is positive if the electric field lines are leaving the surface, and negative if entering the surface. ˆ n  E  Φ  In general, a surface S can be curved and the electric field E  may vary over the surface.We shall be interested in the case where the surface is closed  . A closed surface is asurface which completely encloses a volume. In order to compute the electric flux, wedivide the surface into a large number of infinitesimal area elements ˆ i i i  A ∆ = ∆ An  , asshown in Figure 4.1.3. Note that for a closed surface the unit vector is chosen to pointin the outward  normal direction.ˆ i n   Figure 4.1.3 Electric field passing through an area element i ∆ A  , making an angle θ  withthe normal of the surface.The electric flux through is i ∆ A    cos  E i i i i  E A θ  ∆Φ = ⋅∆ = ∆ EA  (4.1.3)The total flux through the entire surface can be obtained by summing over all the areaelements. Taking the limit and the number of elements to infinity, we have 0 i ∆ → A    0 lim  E iiiS  A d  ∆ → Φ = ⋅ = ⋅ d  ∑∫∫  EAEA         (4.1.4)where the symbol S ∫∫   denotes a double integral over a closed  surface S . In order toevaluate the above integral, we must first specify the surface and then sum over the dotproduct . d  ⋅ EA    4.2   Gauss’s Law Consider a positive point charge Q located at the center o a sphere of radius r  , as shownin Figure 4.2.1. The electric field due to the charge Q isf  r 20 ˆ(/4) Q r  πε  = E  , which points2  in the radial direction. We enclose the charge by an imaginary sphere of radius r  calledthe “Gaussian surface.” Figure 4.2.1 A spherical Gaussian surface enclosing a charge Q .ates, a small surface area element on the sphere is given by (Figure.2.2)In spherical coordin4  2 ˆsin d r d d  θ θ φ  = Ar  (4.2.1) Figure 4.2.2 A small area element on the surface of a sphere of radius r  .Thus, the net electric flux through the area element is ( ) 2   20 4  E  r  0 1sin =sin4 Q Qd d EdA r d d d d  θ θ φ θ θ φ πε  ⎛ ⎞Φ = ⋅ = =⎜ ⎟ EA  (4.2.2)he total flux through the entire surface is πε  ⎝ ⎠  T  20000 4  E S sin = Q Qd d d  π π  θ θ φ πε ε    Φ = ⋅ = ∫∫ ∫ ∫  EA  (4.2.3)us r  has a surface area  The same result can also be obtained by noting that a sphere of radi 2 4  A r  π  = , and since the magnitude of the electric field at any point on the sphericalsurface is 20  /4  E Q r  πε  = , the electric flux through the surface is3
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