Question

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.

Short answer

Inside the cylinder (\(\rho < b\)): $$\mathbf{E}_{in}(\rho)=\frac{a\rho^{3}}{4\epsilon_0}\hat{\rho}$$ Outside the cylinder (\(\rho > b\)): $$\mathbf{E}_{out}(\rho)=\frac{ab^{4}}{4\epsilon_0\rho}\hat{\rho}$$

Step-by-step solution

Find the differential volume element in cylindrical coordinates

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.

Find the total charge enclosed by the Gaussian surface

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: $$Q_{enc}=\pi aL\rho^{4}/4$$

Apply Gauss's Law inside the cylinder

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: $$E_{in}(\rho)=\frac{a\rho^{3}}{4\epsilon_0}$$

Apply Gauss's Law outside the cylinder

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: $$E_{out}(\rho)=\frac{ab^{4}}{4\epsilon_0\rho}$$

Write the final expressions for the electric field

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 (\(\rho > b\)): $$\mathbf{E}_{out}(\rho)=\frac{ab^{4}}{4\epsilon_0\rho}\hat{\rho}$$

Key concepts

Cylindrical Coordinates

Imagine you are positioning yourself along with a number of points around the surface of a cylinder. To describe these points clearly, cylindrical coordinates are the perfect choice.
Unlike Cartesian coordinates which use X, Y, and Z, cylindrical coordinates utilize three parameters: \(\rho\) (the radial distance from the axis), \(\phi\) (the angular displacement from a reference direction), and \(z\) (the height along the axis of the cylinder).
This system is particularly useful when dealing with problems involving circular symmetry, such as our infinitely long cylindrical dielectric.
Think of it as a way to map out positions in a cylindrical-shaped field easily. By knowing these parameters, you can precisely pinpoint any location within the cylindrical structure, which is essential for calculating aspects like volume elements as seen in the given problem.

Gauss's Law

Gauss's Law is a fundamental principle that relates electric fields and charge distributions. It states that the electric flux through a closed surface is equal to the charge enclosed divided by the electric constant, \(\epsilon_0\).

Mathematically, Gauss's Law is written as:
\[\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\epsilon_0}\]
Where \(\mathbf{E}\) is the electric field, and \(d\mathbf{A}\) is the differential area vector on the Gaussian surface.

In our problem, Gauss's Law helps us calculate the electric field inside and outside a cylindrical dielectric by considering a Gaussian surface that is also cylindrical. Gauss's Law ensures that the choice of Gaussian surface simplifies the mathematical calculations, reducing them to manageable integrals.
By strategically applying this law, you can easily determine the electric field environments needed for our solution.

Electric Field Strength

Electric field strength is a measure of the force that would be experienced by a unit positive charge at a given point in an electric field. For our cylindrical dielectric problem, determining the electric field strength involved using the results from Gauss's Law.

Inside the cylinder, the electric field strength, \(\mathbf{E}_{in}(\rho)\), was found to depend on \(\rho\), the distance from the axis of the cylinder. This is described by:
\[\mathbf{E}_{in}(\rho)=\frac{a\rho^{3}}{4\epsilon_0}\hat{\rho}\]
Where \(\hat{\rho}\) is the radial unit vector pointing away from the axis, highlighting the dependency on radial distance within the dielectric.

Outside the cylinder, the electric field changes its dependency based solely on relation to \(\rho\) and the constant factors of the cylinder's properties, leading to:
\[\mathbf{E}_{out}(\rho)=\frac{ab^{4}}{4\epsilon_0\rho}\hat{\rho}\]
Here, the decrease with \(\rho\) indicates how the influence of the dielectric lessens as you move further away.

Dielectrics

Dielectrics are insulating materials that can influence electric fields even though they do not allow current to flow through them.
When considering dielectrics, their ability to be polarized plays a crucial role. In simple terms, when a dielectric is placed in an electric field, it reduces the effective field within it by creating an opposing field.

In our problem, the cylindrical dielectric contains a volumetric charge density, \(\rho_{v}=a \rho^{2}\), where \(a\) is a constant. This distribution influences the electric field formed inside and outside the dielectric.
Understanding the nature of dielectrics helps in appreciating how they cause field lines to bend within the material. This characteristic ensures that potential fields inside the dielectric differ from those outside it, contributing to how we derived the expressions for \(\mathbf{E}_{in}\) and \(\mathbf{E}_{out}\).
By carefully considering these implications of dielectrics, you ensure accurate electric field calculations.