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Chapter 22 Gauss’s Law

Chapter 22 Gauss’s Law. Electric charge and flux (sec. 22.2 & .3) Gauss’s Law (sec. 22.4 & .5) Charges on conductors (sec. 22.6). C 2009 J. F. Becker. A charge inside a box can be probed with a test charge q o to measure E field outside the box.

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Chapter 22 Gauss’s Law

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  1. Chapter 22 Gauss’s Law • Electric charge and flux (sec. 22.2 & .3) • Gauss’s Law (sec. 22.4 & .5) • Charges on conductors (sec. 22.6) C 2009 J. F. Becker

  2. A charge inside a box can be probed with a test charge qo to measure E field outside the box.

  3. The volume (V) flow rate (dV/dt) of fluid through the wire rectangle (a) is vA when the area of the rectangle is perpendicular to the velocity vector v and (b) is vA cosfwhen the rectangle is tilted at an anglef. We will next replace the fluid velocity flow vector v with the electric field vector E to get to the concept of electric fluxFE. Volume flow rate through the wire rectangle.

  4. (a) The electric flux through the surface = EA. (b) When the area vector makes an anglefwith the vector E, the area projected onto a plane oriented perpendicular to the flow is A perp.   = A cosf.  The flux is zero whenf= 90o because the rectangle lies in a plane parallel to the flow and no fluid flows through the rectangle A flat surface in a uniform electric field.

  5. FE=òE . dA =òE dA cosf=ò E dA = Eò dA = E (4p R2) = (1/4peo) q /R2) (4pR2) = q /eo. So we have the electric flux FE= q /eo.  Now we can write Gauss's Law: FE=òE . dA =òEdA cosf =Qencl /eo Electric FLUX through a sphere centered on a point charge q.

  6. The projection of an element of area dA of a sphere of radius R UP onto a concentric sphere of radius 2R. The projection multiplies each linear dimension by 2, so the area element on the larger sphere is 4 dA. The same number of lines of flux pass thru each area element. Flux FEfrom a point charge q.

  7. The projection of the area element dA onto the spherical surface is dA cos f. Flux through an irregular surface.

  8. Spherical Gaussian surfaces around (a) positive and (b) negative point charge.

  9. Gauss’s Law can be used to calculate the magnitude of the E field vector: C 2009 J. F. Becker

  10. Use the following recipe for Gauss’s Law problems:

  11. Use the following recipe for Gauss’s Law problems:1. Carefully draw a figure - location of all charges, direction of electric field vectors E

  12. Use the following recipe for Gauss’s Law problems:1. Carefully draw a figure - location of all charges, direction of electric field vectors E2. Draw an imaginary closed Gaussian surface so that the value of the magnitude of the electric field is constant on the surface and the surface contains the point at which you want to calculate the field. 

  13. Use the following recipe for Gauss’s Law problems:1. Carefully draw a figure - location of all charges, direction of electric field vectors E2. Draw an imaginary closed Gaussian surface so that the value of the magnitude of the electric field is constant on the surface and the surface contains the point at which you want to calculate the field.  3. Write Gauss Law and perform dot product E o dA

  14. Use the following recipe for Gauss’s Law problems:1. Carefully draw a figure - location of all charges, direction of electric field vectors E2. Draw an imaginary closed Gaussian surface so that the value of the magnitude of the electric field is constant on the surface and the surface contains the point at which you want to calculate the field.  3. Write Gauss Law and perform dot product E o dA4. Since you drew the surface in such a way that the magnitude of the E is constant on the surface, you can factor the |E| out of the integral.

  15. Use the following recipe for Gauss’s Law problems:1. Carefully draw a figure - location of all charges, direction of electric field vectors E2. Draw an imaginary closed Gaussian surface so that the value of the magnitude of the electric field is constant on the surface and the surface contains the point at which you want to calculate the field.  3. Write Gauss Law and perform dot product E . dA4. Since you drew the surface in such a way that the magnitude of the E is constant on the surface, you can factor the |E| out of the integral.5. Determine the value of Qencl from your figure and insert it into Gauss's equation.

  16. Use the following recipe for Gauss’s Law problems:1. Carefully draw a figure - location of all charges, direction of electric field vectors E2. Draw an imaginary closed Gaussian surface so that the value of the magnitude of the electric field is constant on the surface and the surface contains the point at which you want to calculate the field.  3. Write Gauss Law and perform dot product E o dA4. Since you drew the surface in such a way that the magnitude of the E is constant on the surface, you can factor the |E| out of the integral.5. Determine the value of Qencl from your figure and insert it into Gauss's equation. 6. Solve the equation for the magnitude of E. C 2009 J. F. Becker

  17. Gaussian surface Under electrostatic conditions, any excess charge resides entirely on the surface of a solid conductor.

  18. Under electrostatic conditions the electric field inside a solid conducting sphere is zero. Outside the sphere the electric field drops off as 1 / r2, as though all the excess charge on the sphere were concentrated at its center. Electric field = zero (electrostatic) inside a solid conducting sphere

  19. A coaxial cylindrical Gaussian surface is used to find the electric field outside an infinitely long charged wire.

  20. A cylindrical Gaussian surface is used to find the electric field of an infinite plane sheet of charge.

  21. Electric field between two oppositely charged parallel plates.

  22. “Volume charge density“: r= charge / unit volume is used to characterize the charge distribution. The electric field of a uniformly charged INSULATING sphere.

  23. The solution of this problem lies in the fact that the electric field inside a conductor is zero and if we place our Gaussian surface inside the conductor (where the field is zero), the charge enclosed must be zero (+ q – q) = 0. Find electric charge q on surface of hole in the charged conductor.

  24. A Gaussian surface drawn inside the conducting material of which the box is made must have zero electric field on it (field inside a con-ductor is zero).  If the Gaussian surface has zero field on it, the charge enclosed must be zero per Gauss's Law. The E field inside a conducting box (a “Faraday cage”) in an electric field.

  25. Review See www.physics.edu/becker/physics51 C 2009 J. F. Becker

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