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www.GOALias.blogspot.comPhysicsFor an area element of a closed surface, ˆn is taken to be the directionof outward normal, by convention.15. Gauss’s law: The flux of electric field through any closed surface S is1/ε 0times the total charge enclosed by S. The law is especially usefulin determining electric field E, when the source distribution has simplesymmetry:(i) Thin infinitely long straight wire of uniform linear charge density λλE =2 πεr0nˆwhere r is the perpendicular distance of the point from the wire andˆn is the radial unit vector in the plane normal to the wire passingthrough the point.(ii) Infinite thin plane sheet of uniform surface charge density σE =σ2 ε0nˆwhere ˆn is a unit vector normal to the plane, outward on either side.(iii) Thin spherical shell of uniform surface charge density σqE = rˆ ( r ≥ R)24 πε0rE = 0 (r < R )where r is the distance of the point from the centre of the shell and Rthe radius of the shell. q is the total charge of the shell: q = 4πR 2 σ.The electric field outside the shell is as though the total charge isconcentrated at the centre. The same result is true for a solid sphereof uniform volume charge density. The field is zero at all points insidethe shellPhysical quantity Symbol Dimensions Unit RemarksVector area element Δ S [L 2 ] m 2 ΔS = ΔS ˆnElectric field E [MLT –3 A –1 ] V m –1Electric flux φ [ML 3 T –3 A –1 ] V m Δφ = E.ΔSDipole moment p [LTA] C m Vector directedfrom negative topositive chargeCharge densitylinear λ [L –1 TA] C m –1 Charge/lengthsurface σ [L –2 TA] C m –2 Charge/area44volume ρ [L –3 TA] C m –3 Charge/volume