Respuesta :
Answer: The electric field is: a) r<a , E0=; b) a<r<b E=ρ (r-a)/εo;
c) r>b E=ρ b (b-a)/r*εo
Explanation: In order to solve this problem we have to use the Gaussian law in diffrengios regions.
As we know,
∫E.dr= Qinside/εo
For r<a --->Qinside=0 then E=0
for a<r<b er have
E*2π*r*L= Q inside/εo in this case Qinside= ρ.Vol=ρ*2*π*r*(r-a)*L
E*2π*r*L =ρ*2*π*r* (r-a)*L/εo
E=ρ*(r-a)/εo
Finally for r>b
E*2π*r*L =ρ*2*π*b* (b-a)*L/εo
E=ρ*b* (b-a)*/r*εo
The electric field is the field, which is surrounded by the electric charged.
The electric field for each case is,
- Case 1- [tex]0
[tex]E=\dfrac{2k_6 \lambda}{r}[/tex]
- Case 2- [tex]a
[tex]E=\dfrac{2k_6 [\lambda+\rho \pi(r^2-a^2)]}{r}[/tex]
- Case 3- [tex]r>b[/tex]
[tex]E=\dfrac{2k_6 [\lambda+\rho \pi(b^2-a^2)]}{r}[/tex]
What is electric field?
The electric field is the field, which is surrounded by the electric charged. The electric field is the electric force per unit charge.
Given information-
The inner radius of the infinitely long cylindrical insulating shell is [tex]a[/tex].
The outer radius of the infinitely long cylindrical insulating shell is [tex]b[/tex].
The uniform volume charge density is [tex]\rho[/tex].
Let [tex]r[/tex] is the radius of the [tex]L[/tex] is the length. Thus,
- Case 1- [tex]0
By the Gauss's law,
[tex]E(2\pi rL)=4\pi k_6 (\lambda L)\\E=\dfrac{2k_6 \lambda}{r}[/tex]
- Case 2- [tex]a
The volume for this case can be given as,
[tex]V=\pi r^2L -\pi a^2 L\\V=\pi L(r^2-a^2)[/tex]
Put this values in the Gauss's law,
[tex]E(2\pi rL)=4\pi k_6 (\lambda L+\rho V)\\E(2\pi rL)=4\pi k_6 (\lambda L+\rho \pi L(r^2-a^2))\\E=\dfrac{2k_6 [\lambda+\rho \pi(r^2-a^2)]}{r}[/tex]
- Case 3- [tex]r>b[/tex]
The volume for this case can be given as,
[tex]V=\pi b^2L -\pi a^2 L\\V=\pi L(b^2-a^2)[/tex]
Put this values in the Gauss's law,
[tex]E(2\pi rL)=4\pi k_6 (\lambda L+\rho V)\\E(2\pi rL)=4\pi k_6 (\lambda L+\rho \pi L(b^2-a^2))\\E=\dfrac{2k_6 [\lambda+\rho \pi(b^2-a^2)]}{r}[/tex]
Thus the electric field for each case is,
- Case 1- [tex]0
[tex]E=\dfrac{2k_6 \lambda}{r}[/tex]
- Case 2- [tex]a
[tex]E=\dfrac{2k_6 [\lambda+\rho \pi(r^2-a^2)]}{r}[/tex]
- Case 3- [tex]r>b[/tex]
[tex]E=\dfrac{2k_6 [\lambda+\rho \pi(b^2-a^2)]}{r}[/tex]
Learn more about electric field here;
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