Respuesta :
Answer:
1) The resistance of the glass rod is 2122.1 Ω
2) The current density is [tex]\frac{12\sigma_0 \Delta V }{L}[/tex]
3) the steady-state magnitude of the electric field E is [tex]= \frac{\Delta V\cdot12}{L^3} x^2i[/tex].
Explanation:
Here we note that
the conductivity varies with position x measured from the center of the rod as
σ(x) = σ₀L²/x² with σ₀ = 4 × 10⁻² (Ω·m⁻¹)
the resistance of a small length x is given by
dR = [tex]\frac{\rho dx}{A} = \frac{dx}{\sigma _c A} = \frac{4x^{2}dx }{\sigma _0 L^2\pi d^2 }[/tex]
Where:
ρ = Resistivity
From the center, we integrate over the entire length of the cable as follows
[tex]R = \int\limits^{\frac{L}{2}}_{-\frac{L}{2}} {} \, dR = \int\limits^{\frac{L}{2}}_{-\frac{L}{2}} {} \, \frac{4x^{2}dx }{\sigma _0 L^2\pi d^2 } = \frac{4x^{3}dx }{3\sigma _0 L^2\pi d^2}|_{-L/2}^{L/2} =\frac{L}{3\sigma_0 \pi d^2 }[/tex]
Which gives
R = [tex]\frac{0.02 }{3\times 0.04\times \pi \times0.05^2}[/tex] = 2122.066 Ω
2) The current density [tex]|\overrightarrow{\rm J } (x)|[/tex] = [tex]\frac{I}{A}[/tex]
Where:
I = Current = [tex]\frac{Pd}{R}[/tex]and pd = ΔV
A = Area = πd²/4
Therefore current desity = [tex]\frac{4\Delta V}{R\pi d^2}[/tex]
Substituting = [tex]R =\frac{L}{3\sigma_0 \pi d^2 }[/tex]
Then we have
Current density = [tex]\frac{12\sigma_0 \Delta V }{L}[/tex]
3) The steady-state magnitude of the electric field is given by
Magnitude of the Electric field
[tex]\overrightarrow{\rm E}(x) = \frac{\overrightarrow{\rm J } (x)}{\sigma} = \frac{12\sigma_0 \Delta V /L}{\sigma_0 L^2/x^2}i = \frac{\Delta V\cdot12}{L^3} x^2i[/tex]
