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SVIT ,VASAD

SVIT ,VASAD. EEE PRESENTATION. TOPIC:GAUSS’S LAW Presentated by: VIVEK CHAUHAN- 13BEECG010 BHAVIN PANDYA- 13BEEN022 DUTT PAREKH - 13BEECG002 YASH HINGU - 13BEECG009. Gauss’s Law. Electric charge and flux. Gauss’s Law. Charges on conductors.

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SVIT ,VASAD

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  1. SVIT,VASAD

  2. EEE PRESENTATION TOPIC:GAUSS’S LAW Presentated by: VIVEK CHAUHAN- 13BEECG010 BHAVIN PANDYA- 13BEEN022 DUTT PAREKH - 13BEECG002 YASH HINGU - 13BEECG009

  3. Gauss’s Law • Electric charge and flux. • Gauss’s Law. • Charges on conductors.

  4. learning Goals: • How you can determine the amount of charge within a closed surface by examining the electric field on the surface!• What is meant by electric flux and how you can calculate it.• How to use Gauss’s Law to calculate the electric field due to a symmetric distribution of charges.

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

  6. 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.

  7. (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.

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

  9. 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.

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

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

  12. Gauss’s Law can be used to calculate the magnitude of the E field vector: C 2012 J. F. Becker

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

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

  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. 

  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 dA

  17. 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.

  18. 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.

  19. Use the following recipe for Gauss’s Law problems:1. Carefully draw a figure - location of all charges, and 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.

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

  21. 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

  22. A coaxial cylindricalGaussian surface is used to find the electric field outside an infinitely long charged wire.

  23. A cylindricalGaussian surface is used to find the electric field of an infinite plane sheet of charge.

  24. (Ignoring edge effects) Electric field between two (large) oppositely charged parallel plates.

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

  26. +q 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 the electric charge q on surface ofa hole in the charged conductor.

  27. 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. E The E field inside a conducting box (a “Faraday cage”) in an electric field.

  28. THANK YOU

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