Electrical Field on a Gaussian Surface

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• E-field of a spherical shell with uniform volume charge

  Considering a spherical Gaussian surface with radius r, the electric field is parallel to dA vector everywhere on the surface, so the dot product becomes a simple multiplication because the angle between E vector and the dA vector is zero everywhere.

• Gauss's Law.

  The symmetry of the problem suggests that if we use a concentric cylindrical Gauss surface, the electrical field will be perpendicular to the surface envelope, hence: (1.2) And the magnitude of the electric field is constant on the surface: (

• Two problems on electrostatics

  Applying Gauss' theorem at the Gaussian surface we get : ∫E.ds = Q/Єo where E = Electric field on the Gaussian surface, Q = Total charge enclosed by the Gaussian surface (the charge on the inner shell + the charge on the outer shell) E x 4Π x

• Gauss Law Infinite Plans of Surface Charge Densities

  Let the field strength outside the plate system be E in magnitude and away from the system, than the value of for each plane surface of the Gaussian surface will be E*A and thus the total value of for the whole Gaussian surface will be 2*E*A.

• What is the magnitude of electric field near the center of the sheet?

  THe total charge enclosed by our gaussian surrface is sigma*A where sigma is the surface charge density.

• Electrostatics: Surface charge density and electric field

  For spherically symmetric field the surface integral is given by E*4r2 hence (i) if r < a charge within the Gaussian surface is +q E*4r2 = q/ Or (ii) if q < r < R the net charge within the Gaussian surface is +q -q = 0 hence E*4r2

• Application of Gauss' law

  Q = 2Πρ0(RA)3 Electric field intensity E at any point on the Gaussian surface will be same in magnitude (as all points on the surface are equidistant from the centre of the charged sphere) and the direction of the E vector at any point on the surface

• Use Gauss's Law to find the electic field, electric potential

  Then the charge enclosed by the Gaussian surface is: Q_in = pV' where V' = (4/3)*pi*R^3 Since the magnitude of the electric field is constant everywhere on the spherical Gaussian surface and is normal to the surface at each point, gauss' law gives

• application of Gauss law to find the electric field

  (b) The net charge per unit length inside the enclosing Gaussian surface is the total charge on the rod and the shell. Thank you for using Brainmass.