A long cylinder (Fig. 2.21) carries a charge density that is proportional to the distance from the axis: ρ = ks, for some constant k. Find the electric field inside this cylinder.

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Accepted Answer

Draw a Gaussian cylinder of length l and radius s. For this surface, Gauss’s law states:

[latex]\oint\limits_{S}^{}{E.da} =\frac{1}{\epsilon _{0}} Q_{enc.}[/latex]

The enclosed charge is

[latex] Q_{enc}=\int{ ho d au }=\int{(k\acute{s} )(\acute{s}d\acute{s}d\phi dz)}=2\pi kl\int_{0}^{s}{\acute{s}^{2}d\acute{s}}=\frac{2}{3}\pi kls^{3} [/latex]

(I used the volume element appropriate to cylindrical coordinates, Eq.1 .78,

[latex]d au =sdsd\phi dz[/latex] (1.78)

and integrated φ from 0 to 2π, dz from 0 to l. I put a prime on the integration variable [latex]\acute{s}[/latex], to distinguish it from the radius s of the Gaussian surface.)
Now, symmetry dictates that E must point radially outward, so for the curvedportion of the Gaussian cylinder we have:

[latex]\int{E.da}=\int{\left|E ight|da }=\left|E ight|\int{da} =\left|E ight|2\pi sl, [/latex]

while the two ends contribute nothing (here E is perpendicular to da). Thus,

[latex]\left|E ight|2\pi sl=\frac{1}{\epsilon _{0}}\frac{2}{3}\pi kls^{3} , [/latex]

or, finally,

[latex]E=\frac{1}{3\epsilon _{0}}ks^{2}\hat{s} [/latex]

Question 2.4: A long cylinder (Fig. 2.21) carries a charge density that is...

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