Respuesta :
a)
i) Potential for r < a: [tex]V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})[/tex]
ii) Potential for a < r < b: [tex]V(r)=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}[/tex]
iii) Potential for r > b: [tex]V(r)=0[/tex]
b) Potential difference between the two cylinders: [tex]V_{ab}=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})[/tex]
c) Electric field between the two cylinders: [tex]E=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}[/tex]
Explanation:
a)
Here we want to calculate the potential for r < a.
Before calculating the potential, we have to keep in mind that the electric field outside an infinite wire or an infinite cylinder uniformly charged is
[tex]E=\frac{\lambda}{2\pi \epsilon_0 r}[/tex]
where
[tex]\lambda[/tex] is the linear charge density
r is the distance from the wire/surface of the cylinder
By integration, we find an expression for the electric potential at a distance of r:
[tex]V(r) =\int Edr = \frac{\lambda}{2\pi \epsilon_0} ln(r)[/tex]
Inside the cylinder, however, the electric field is zero, because the charge contained by the Gaussian surface is zero:
[tex]E=0[/tex]
So the potential where the electric field is zero is constant:
[tex]V=const.[/tex]
iii) We start by evaluating the potential in the region r > b. Here, the net electric field is zero, because the Gaussian surface of radius r here contains a positive charge density [tex]+\lambda[/tex] and an equal negative charge density [tex]-\lambda[/tex]. Therefore, the net charge is zero, so the electric field is zero.
This means that the electric potential is constant, so we can write:
[tex]\Delta V= V(r) - V(b) = 0\\\rightarrow V(r)=V(b)[/tex]
However, we know that the potential at b is zero, so
[tex]V(r)=V(b)=0[/tex]
ii) The electric field in the region a < r < b instead it is given only by the positive charge [tex]+\lambda[/tex] distributed over the surface of the inner cylinder of radius a, therefore it is
[tex]E=\frac{\lambda}{2\pi r \epsilon_0}[/tex]
And so the potential in this region is given by:
[tex]V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}[/tex] (1)
i) Finally, the electric field in the region r < a is zero, because the charge contained in this region is zero (we are inside the surface of the inner cylinder of radius a):
E = 0
This means that the potential in this region remains constant, and it is equal to the potential at the surface of the inner cylinder, so calculated at r = a, which can be calculated by substituting r = a into expression (1):
[tex]V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})[/tex]
And so, for r<a,
[tex]V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})[/tex]
b)
Here we want to calculate the potential difference between the surface of the inner cylinder and the surface of the outer cylinder.
We have:
- Potential at the surface of the inner cylinder:
[tex]V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})[/tex]
- Potential at the surface of the outer cylinder:
[tex]V(b)=0[/tex]
Therefore, the potential difference is simply equal to
[tex]V_{ab}=V(a)-V(b)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})[/tex]
c)
Here we want to find the magnitude of the electric field between the two cylinders.
The expression for the electric potential between the cylinders is
[tex]V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}[/tex]
The electric field is just the derivative of the electric potential:
[tex]E=-\frac{dV}{dr}[/tex]
so we can find it by integrating the expression for the electric potential. We find:
[tex]E=-\frac{d}{dr}(\frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}[/tex]
So, this is the expression of the electric field between the two cylinders.
Learn more about electric fields:
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