at every point, and its magnitude is constant, since it depends only on r . The surface area of the curved part is π rl , where l is the length of the cylinder. FIGURE . (a) Electric field due to an infinitely long thin straight wire is radial, (b) The Gaussian surface for a long thin wire of uniform linear charge density.
Flux through the Gaussian surface flux through the curved cylindrical part of the surface E × π rl The surface includes charge equal to λ l . Gauss’s law then gives E × π rl = λ l / ε i.e., E = λ Vectorially, E at any point is given by ˆ λ E n ( . ) where ˆ n is the radial unit vector in the plane normal to the wire passing through the point. E is directed outward if λ is positive and inward if λ is negative.
Note that when we write a vector A as a scalar multiplied by a unit vector, i.e., as A = A ˆ a , the scalar A is an algebraic number. It can be negative or positive. The direction of A will be the same as that of the unit vector ˆ a if A > and opposite to ˆ a if A < . When we want to restrict to non-negative values, we use the symbol A and call it the modulus of A .
Thus, ≥ A . Also note that though only the charge enclosed by the surface ( λ l ) was included above, the electric field E is due to the charge on the entire wire. Further, the assumption that the wire is infinitely long is crucial. Without this assumption, we cannot take E to be normal to the curved part of the cylindrical Gaussian surface.
However, Eq. ( . ) is approximately true for electric field around the central portions of a long wire, where the end effects may be ignored. .
. Field due to a uniformly charged infinite plane sheet Let σ be the uniform surface