Applications Of Gauss's Law

Use Gauss's Law to derive electric field formulas for symmetric charge distributions (wire, sheet, shell).

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Concept

Field of an Infinitely Long Wire

Derivation using a cylindrical Gaussian surface.

Imagine an infinitely long, thin straight wire with a uniform linear charge density λ\lambda. Because the wire is perfectly straight and infinite, it has cylindrical symmetry. This means the electric field must look the same from any angle around the wire and must point radially outward (if λ>0\lambda > 0).

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Concept

Field of a Plane Sheet and Spherical Shell

Derivations for a sheet and a thin shell.

Consider an infinite flat sheet with a uniform surface charge density σ\sigma. By symmetry, the electric field E\vec{E} must point straight away from the sheet (if positive) and cannot depend on your yy or zz coordinates along the plane.

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Gauss's Law Applications Summary

Matrix of formulas for wire, sheet, and shell.

Point Charge
Spherical symmetry
Shape
Around a point
Scaling
Infinite Wire
Cylindrical symmetry
Shape
Around a line
Scaling
Infinite Sheet
Planar symmetry
Shape
Near a plane
Scaling
Constant, no dependence
Spherical Shell
Shell symmetry
Outside
Inside
Outside
Acts like a point charge
Inside
Electric field is zero
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Worked Example

Early Atom Model

Derives field for a uniform spherical charge distribution.

Problem

An early model for an atom considered it to have a positively charged point nucleus of charge ZeZe, surrounded by a uniform density of negative charge up to a radius RR. The atom as a whole is neutral. For this model, what is the electric field at a distance rr from the nucleus?

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Faded Example: Field of a Sheet

Calculate linear charge density given the field.

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Problem: An infinite line charge produces an electric field of 9×104 N/C9 \times 10^4 \text{ N/C} at a distance of 2 cm2 \text{ cm}. Calculate the linear charge density λ\lambda.

Solution: We know the formula for the electric field of an infinite line charge is E=λ2πϵ0rE = \frac{\lambda}{2 \pi \epsilon_0 r}. To find the linear charge density, we rearrange the formula to isolate λ\lambda: λ=\lambda = . Now, we substitute the given values: E=9×104 N/CE = 9 \times 10^4 \text{ N/C} and r=0.02 mr = 0.02 \text{ m}. Using the electrostatic constant 14πϵ0=9×109 N m2/C2\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2, we can write 2πϵ0=118×1092 \pi \epsilon_0 = \frac{1}{18 \times 10^9}. Plugging these into our rearranged equation gives λ=9×104×0.0218×109\lambda = \frac{9 \times 10^4 \times 0.02}{18 \times 10^9}. Calculating this yields the final linear charge density: λ=\lambda =  C/m\text{ C/m}.

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quiz

Distance Dependence Match

MCQ testing the r-dependence of different fields.

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How does the magnitude of the electric field of an infinitely long, uniformly charged wire scale with the radial distance rr?