An infinitely long cylindrical dielectric of radius \(b\) contains charge within its volume of density \(\rho_{v}=a \rho^{2}\), where \(a\) is a constant. Find the electric field strength, \(\mathbf{E}\), both inside and outside the cylinder.

In cylindrical coordinates, we can express the differential volume element as \(dV=\rho d\rho d\phi dz\). We will use this to determine the total charge enclosed by a Gaussian surface within the cylinder.

Consider a Gaussian cylinder of radius \(\rho\) and length L centered on the axis of the dielectric cylinder. For the region inside the dielectric, the total charge enclosed can be found by integrating the charge density over the Gaussian cylinder's volume: $$Q_{enc}=\int_0^L\int_0^{2\pi}\int_0^\rho (a\rho'^{2})\rho' d\rho'd\phi'dz'$$ Solving this integral, we get: $$Q_{enc}=\pi aL\rho^{4}/4$$

Applying Gauss's Law to the Gaussian cylinder: $$\oint\mathbf{E}_{in}\cdot d\mathbf{A}=\frac{Q_{enc}}{\epsilon_0}$$ The electric field is symmetric about the axis and only depends on \(\rho\). Thus, \(\mathbf{E}_{in} = E_{in}(\rho)\hat{\rho}\). The area vector, \(d\mathbf{A}\), is also in the radial direction: \(d\mathbf{A}=\rho d\phi dz \hat{\rho}\). Therefore, the dot product simplifies to: $$\oint E_{in}(\rho) \rho d\phi dz = \frac{\pi aL\rho^{4}}{4\epsilon_0}$$ Integrating over the Gaussian surface and solving for the electric field inside the cylinder, we get: $$E_{in}(\rho)=\frac{a\rho^{3}}{4\epsilon_0}$$

For the region outside the dielectric (i.e., \(\rho > b\)), the Gaussian cylinder encloses the whole dielectric cylinder. Thus, the total charge enclosed can be found by integrating the charge density over the entire dielectric cylinder's volume: $$Q_{enc,tot}=\int_0^L\int_0^{2\pi}\int_0^b (a\rho'^{2})\rho' d\rho'd\phi'dz'$$ Solving this integral, we get: $$Q_{enc,tot}=\pi aLb^{4}/4$$ Applying Gauss's Law to the Gaussian cylinder: $$\oint\mathbf{E}_{out}\cdot d\mathbf{A}=\frac{Q_{enc,tot}}{\epsilon_0}$$ Similar to the inside case, \(\mathbf{E}_{out} = E_{out}(\rho)\hat{\rho}\), and \(d\mathbf{A}=\rho d\phi dz \hat{\rho}\). Therefore, the dot product simplifies to: $$\oint E_{out}(\rho) \rho d\phi dz = \frac{\pi aLb^{4}}{4\epsilon_0}$$ Integrating over the Gaussian surface and solving for the electric field outside the cylinder, we get: $$E_{out}(\rho)=\frac{ab^{4}}{4\epsilon_0\rho}$$

Now we can summarize the electric field strength both inside and outside the cylindrical dielectric: Inside the cylinder (\(\rho < b\)): $$\mathbf{E}_{in}(\rho)=\frac{a\rho^{3}}{4\epsilon_0}\hat{\rho}$$ Outside the cylinder (\(\rho > b\)): $$\mathbf{E}_{out}(\rho)=\frac{ab^{4}}{4\epsilon_0\rho}\hat{\rho}$$