Problem with electric flux density

bibo_dvd · Jun 19, 2014

hello guys :)
i need your help with this problem

i can understand that in (a) we will use the integral of row(v) * dv to get the charge
and in (b) we will use the formule D(r)=Q/[4*PI*r^2]

but in (C) should it be zero or what ??
and if it should be zero why in the solution manual it uses the formula of D(r)=Q[4*PI*(r^2)] again ?!

i don't know what is the right answer for this ..so i need your help ..Thanks guys :)

Textbook problem 3.9: uniform volume charge density of 80 μC/m³ for 8 mm < r < 10 mm, with ρv = 0 for 0 < r < 8 mm; parts (a), (b) and (c) ask for total charge, D_r at r = 10 mm, and D_r at r = 20 mm.

Typeset solution to part c: D_r(20 mm) = 164 x 10^-12 / (4π(.02)^2) = 3.25 x 10^-8 C/m^2 = 32.5 nC/m^2, with the final value underlined.

jackarms · Jun 19, 2014

No, the answer you have -- ##32.5 \frac{nC}{m^{2}}## -- seems right. The only change in part c is that you're looking farther away from the charge distribution, so there's still going to be electric field there and so the flux density will be nonzero, but since you're farther away from the charge, the density is going to be less.

rude man · Jun 19, 2014

The Gaussian surface area A at r = 20 mm is what multiple of the surface at r = 10 mm? Use that to determine D(20mm)*A(20mm) = D(10mm)*(A(10mm).

Problem with electric flux density

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