An electric field calculation for a linear charge using Gauss’s Law determines the electric field (
) around an infinitely long, straight line of uniform charge density. Gauss’s Law states that the net electric flux through any closed surface equals the enclosed charge divided by the permittivity of free space ( ε0epsilon sub 0
). By exploiting the cylindrical symmetry of a line charge, this model simplifies a complex three-dimensional integration into a straightforward algebraic calculation.
The resulting formula for the electric field at a perpendicular distance from an infinite line charge is:
E=λ2πε0rcap E equals the fraction with numerator lambda and denominator 2 pi epsilon sub 0 r end-fraction
(lambda) represents the linear charge density (charge per unit length). The Step-by-Step Gauss’s Law Derivation
To calculate the electric field, we follow a structured four-step process based on spatial symmetry. 1. Choose the Gaussian Surface
We select an imaginary, closed cylindrical Gaussian surface of radius and length
, coaxial with the line charge. This shape matches the cylindrical symmetry of the electric field, which radiates straight outward radially from the line. 2. Evaluate the Enclosed Charge ( Qenccap Q sub enc end-sub
The total charge trapped inside our closed cylinder depends only on the length of the cylinder and how densely packed the charges are: Qenc=λLcap Q sub enc end-sub equals lambda cap L 3. Calculate the Electric Flux ( ΦEcap phi sub cap E
Electric flux measures the electric field passing through the surface. The cylinder has three distinct faces:
The Top and Bottom Caps: The electric field lines run parallel to these flat surfaces. No field lines pierce through them, meaning the flux through both caps is zero.
The Curved Side: The electric field lines are perfectly perpendicular to this curved surface everywhere. The field strength is also constant at a fixed radius
is uniform, we multiply the electric field by the surface area of the curved cylinder side (
ΦE=∮E⋅dA=E(2πrL)cap phi sub cap E equals contour integral of cap E center dot d cap A equals cap E open paren 2 pi r cap L close paren 4. Apply Gauss’s Law to Solve for
We substitute our flux and enclosed charge into the Gauss’s Law equation (
E(2πrL)=λLε0cap E open paren 2 pi r cap L close paren equals the fraction with numerator lambda cap L and denominator epsilon sub 0 end-fraction Notice that the arbitrary length appears on both sides and cancels out completely. Isolating gives the final model equation:
E=λ2πε0rcap E equals the fraction with numerator lambda and denominator 2 pi epsilon sub 0 r end-fraction Key Properties of the Model Inverse
Relationship: Unlike a point charge where the field drops off sharply with distance squared (
1r2the fraction with numerator 1 and denominator r squared end-fraction
), a linear charge field decreases much more slowly, dropping off inversely with distance ( 1r1 over r end-fraction Directionality: If the line charge density is positive (
), the electric field vectors point radially outward. If negative ( ), they point radially inward.
The Infinite Assumption: This exact mathematical model assumes an infinitely long wire. However, it serves as a highly accurate approximation for real-world finite wires, provided the distance
is very small compared to the total length of the wire, away from the wire ends. Visualizing the Field Decay
The curve below illustrates how the electric field strength rapidly decays as you move radially away from the linear axis of charge. If you want to practice or explore further,
Compare this model directly to Gauss’s Law models for planar or spherical charges.
Look at what happens to the math when the wire has a finite length.
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