Gauss’s Law

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I this presentation me and group had worked on Guass's Law. It includes Introduction, Statements, Understandings and Applications in detail..

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Gauss’s Law

  1. 1.  Gauss’s law was formulated by German scientist Carl Friedrich Gauss in 1835, but was not published until 1867.  Gauss's law, also known as Gauss's flux theorem and Maxwell’s first equation.  Gauss’s law is relating the distribution of electric charge to the resulting electric field.  Gauss’ Law allows us to find electric fields without needing to integrate  The electric field of a given charge distribution can in principle be calculated using Gauss’ law. Carl Friedrich
  2. 2. The total electric flux through a closed surface is proportional to the enclosed charge Where: E = Electric Field dA = Area Vector Σq = Sum of all charges ε0 = Permittivity of free space Value of Permittivity of free space ε0=8.8542x10-12 C2/(N m2)
  3. 3.  Gauss’s Law is just a flux calculation  Gauss’s Law only applies to closed surfaces called Gaussian Surface.  Gauss’s Law directly relates electric flux to the charge distribution that creates it.  Gauss's law can be used to derive Coulomb's law, and vice versa.
  4. 4. Electric Field Intensity due to Spherical Charge Distribution Consider a charge “q” uniformly distributed on a closed sphere of radius “a” as shown in figure
  5. 5. When point “p” Lies Outside the Charge Sphere  E . dA = q / e0  E . dA = E dA = E A  A = 4 r2  E A = E 4 r2 = q / e0  k = 1 / 4 0  0 = permittivity  0 = 8.85x10-12 C2/Nm2  2 04 1 r q E
  6. 6. When point “p” Lies Inside the Charge Sphere From figure it is clear that there is no charge enclosed by Gaussian surface therefore flux passes through it is zero, it means electric field intensity is also zero.  E 4 r2 = q / e0  For Gaussian surface q=0  E 4 r2 = 0 / e0  E = 0 / 4 r2  E = 0
  7. 7. When point “p” Lies on the Surface of Charge Sphere  E 4 r2 = q / e0  E 4 a2 = q / e0 (where r = a)  σ= q / ∆A  q = σ . ∆A  q= σ . 4 a2  E 4 a2 = σ . 4 a2 / e0  E = σ / e0
  8. 8. Electric Field Intensity due to Infinite Sheet of Charge  Select Gauss surface In this case a cylindrical pillbox  Calculate the flu of the electric field through the Gauss surface = 2 E A  Equate = qencl/ 0 2EA = qencl/ 0  Solve for E: E = qencl / 2 A 0 = / 2 0 (with = qencl / A)
  9. 9. Electric Field Intensity due to Oppositely Charge Sheet  E = E1 + E2  E = σ / 2 e0 + σ / 2 e0  E = 2 σ / 2 e0  E = σ / e0

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