Along coaxial cable consists of an inner cylindrical conductor with ra

a) To calculate the electric field at any point between the cylinders, a distance $r$ from the axis, we can make use of Gauss's law.
Consider a cylindrical Gaussian surface of radius $r$, coaxial with the cylinders. The electric field is radially directed and has the same magnitude at every point on this surface. Thus, the electric field is constant along the surface.
By Gauss's law, the electric flux through the Gaussian surface is given by:
$\Phi =\frac{Q_{\text{enc}}}{{\epsilon }_0}$
Since the system has cylindrical symmetry, the electric field is perpendicular to the surface, and the flux can be simplified as:
$\Phi =E·2\pi rL,$
where $L$ is the length of the Gaussian surface. $Q_{\text{enc}}$ represents the charge enclosed by the Gaussian surface.
The charge enclosed by the Gaussian surface is the charge per unit length $\lambda$ multiplied by the length of the Gaussian surface $L$. Therefore, $Q_{\text{enc}}=\lambda L$.
Plugging these values into the equation for electric flux, we have:
$E·2\pi rL=\frac{\lambda L}{{\epsilon }_0}.$
Simplifying, we find the electric field at any point between the cylinders:
$E=\frac{\lambda }{2\pi {\epsilon }_{0}r}.$
b) To calculate the electric field at any point outside the outer cylinder, we can again use Gauss's law. We consider a Gaussian surface outside the outer cylinder. Since the outer cylinder has no net charge, there is no charge enclosed by the Gaussian surface, and therefore the electric flux through the surface is zero.
According to Gauss's law, the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space (${\epsilon }_0$). Since there is no charge enclosed, the electric flux is zero.
Since the electric field is perpendicular to the surface, and the flux is zero, we conclude that the electric field at any point outside the outer cylinder is also zero.
c) To graph the magnitude of the electric field as a function of the distance $r$ from the axis of the cable, from $r=0$ to $r=2c$, we can substitute the values of $\lambda$, $a$, $b$, $c$, and ${\epsilon }_0$ into the equation we derived in part (a):
$E=\frac{\lambda }{2\pi {\epsilon }_{0}r}.$
The graph will have $r$ on the x-axis and $E$ on the y-axis, ranging from $r=0$ to $r=2c$. Note that the value of $E$ will depend on the specific values of $\lambda$, $a$, $b$, $c$, and ${\epsilon }_0$ in the problem.
d) To find the charge per unit length on the inner surface and the outer surface of the outer cylinder, we need to consider the total charge enclosed by the surfaces.
The inner surface of the outer cylinder encloses the charge carried by the inner conductor. Therefore, the charge per unit length on the inner surface of the outer cylinder is $\lambda$.
The outer surface of the outer cylinder does not enclose any net charge since the outer cylinder has no net charge. Therefore, the charge per unit length on the outer surface of the outer cylinder is zero.

Step 1:
a) The electric field at any point between the cylinders, a distance $r$ from the axis, can be calculated using Gauss's law. Since the electric field is radially symmetric, we can consider a cylindrical Gaussian surface of radius $r$ between the cylinders.
The electric field $𝐄$ is directed radially outward and has a magnitude $E$ at every point on the Gaussian surface. The Gaussian surface encloses only the inner cylinder.
Applying Gauss's law, we have:
$\oint 𝐄·d𝐀=\frac{Q_{\text{enc}}}{{ϵ}_0}$
Since the electric field is constant in magnitude and perpendicular to the Gaussian surface, the left-hand side of the equation simplifies to $E·2\pi r·L$, where $L$ is the length of the Gaussian surface.
The charge enclosed $Q_{\text{enc}}$ is equal to the charge per unit length $\lambda$ multiplied by the length $L$ of the Gaussian surface.
Therefore, we have:
$E·2\pi r·L=\frac{\lambda L}{{ϵ}_0}$
Simplifying the equation gives:
$E=\frac{\lambda }{2\pi {ϵ}_{0}r}$
Hence, the electric field at any point between the cylinders, a distance $r$ from the axis, is given by:
$\boxed{E=\frac{\lambda }{2\pi {ϵ}_{0}r}}$
Step 2:
b) The electric field at any point outside the outer cylinder can be calculated in a similar manner. We consider a cylindrical Gaussian surface of radius $r$ outside the outer cylinder. The Gaussian surface encloses both the inner and outer cylinders.
Using Gauss's law again, we have:
$\oint 𝐄·d𝐀=\frac{Q_{\text{enc}}}{{ϵ}_0}$
The left-hand side of the equation is now the sum of the electric field contributions from both cylinders. Since the electric field is directed radially outward, the contribution from the outer cylinder is zero since it has no net charge.
The contribution from the inner cylinder can be calculated in a similar manner as in part (a), considering a length $L$ of the inner cylinder enclosed by the Gaussian surface.
Therefore, we have:
$E·2\pi r·L=\frac{\lambda L}{{ϵ}_0}$
Simplifying the equation gives:
$E=\frac{\lambda }{2\pi {ϵ}_{0}r}$
So, the electric field at any point outside the outer cylinder is the same as the electric field between the cylinders and is given by:
$\boxed{E=\frac{\lambda }{2\pi {ϵ}_{0}r}}$
Step 3:
c) To graph the magnitude of the electric field as a function of the distance $r$ from the axis of the cable, from $r=0$ to $r=2c$, we can substitute the values of $r$ into the equation obtained in parts (a) and (b). The magnitude of the electric field will be the same for both regions.
$E=\frac{\lambda }{2\pi {ϵ}_{0}r}$
We can plot this equation as a function of $r$ in the given range.
Step 4:
d) To find the charge per unit length on the inner surface of the outer cylinder, we need to consider the charge enclosed by a Gaussian surface just inside the outer cylinder.
Since the outer cylinder has no net charge, the charge per unit length on its inner surface is zero. Therefore, the charge per unit length on the inner surface of the outer cylinder is given by:
$\boxed{{\lambda }_{\text{inner}}=0}$
To find the charge per unit length on the outer surface of the outer cylinder, we need to consider the charge enclosed by a Gaussian surface just outside the outer cylinder.
Since the outer cylinder has an inner radius $b$ and an outer radius $c$, the charge enclosed by the Gaussian surface is the charge per unit length $\lambda$ multiplied by the length $L$ of the Gaussian surface.
Therefore, we have:
$Q_{\text{enc}}=\lambda L$
Applying Gauss's law, we have:
$\oint 𝐄·d𝐀=\frac{Q_{\text{enc}}}{{ϵ}_0}$
Since the electric field is constant in magnitude and directed radially outward, the left-hand side of the equation simplifies to $E·2\pi r·L$, where $r$ is the radius of the Gaussian surface.
Substituting the values, we get:
$E·2\pi c·L=\frac{\lambda L}{{ϵ}_0}$
Simplifying the equation gives:
$E=\frac{\lambda }{2\pi {ϵ}_{0}c}$
Since the electric field on the outer surface of the outer cylinder is equal to the magnitude of the electric field between the cylinders, we can use the same equation obtained in parts (a) and (b).
Therefore, the charge per unit length on the outer surface of the outer cylinder is given by:
$\boxed{{\lambda }_{\text{outer}}=\frac{\lambda }{2\pi {ϵ}_{0}c}}$
In summary:
- The charge per unit length on the inner surface of the outer cylinder is zero (${\lambda }_{\text{inner}}=0$).
- The charge per unit length on the outer surface of the outer cylinder is ${\lambda }_{\text{outer}}=\frac{\lambda }{2\pi {ϵ}_{0}c}$.