Question

Volume charge density is located as follows: \(\rho_{v}=0\) for \(\rho<1 \mathrm{~mm}\) and for \(\rho>2 \mathrm{~mm}, \rho_{v}=4 \rho \mu \mathrm{C} / \mathrm{m}^{3}\) for \(1<\rho<2 \mathrm{~mm} .\) (a) Calculate the total charge in the region \(0<\rho<\rho_{1}, 0

Short answer

To summarize the solutions of each part: a) The total charge in the region is given by: $$ Q = 2\pi L\cdot \frac{4\mu C/m^3(\rho_1^3 - 1)}{3}. $$ b) The electric displacement \(D_\rho\) at \(\rho = \rho_1\) is given by: $$ D_\rho = \frac{4\mu C/m^3(\rho_1^3 - 1)}{3\rho_1}. $$ c) The values of \(D_\rho\) at the specified values of \(\rho\) are: - \(\rho = 0.8\) mm: \(D_\rho = 0 \mu C/m^2\) - \(\rho = 1.6\) mm: \(D_\rho \approx 1.409 \mu C/m^2\) - \(\rho = 2.4\) mm: \(D_\rho \approx 1.667 \mu C/m^2\).

Step-by-step solution

Identify volume charge density function

The volume charge density \(\rho_v\) is given as follows: $$\rho_{v}=0 \text{ for } \rho<1 \mathrm{~mm} \text{ and for } \rho>2 \mathrm{~mm},$$ $$\rho_{v}=4 \rho \mu \mathrm{C} / \mathrm{m}^{3} \text{ for } 1<\rho<2 \mathrm{~mm}.$$

Calculate the total charge

The total charge in the region is given by integrating the volume charge density over the volume under consideration: $$Q = \int_{0}^{2\pi} \int_{0}^{L} \int_{0}^{\rho_1} \rho_v \cdot \rho d\rho dz d\phi. $$ Since \(\rho_v = 0\) for \(\rho < 1\) mm, we can consider only the region where \(\rho_v = 4\rho \mu C / m^3\). Thus, the integral will give: $$Q = \int_{0}^{2\pi} d\phi \int_{0}^{L} dz \int_{1}^{\rho_1} (4 \rho \mu C/m^3) \rho d\rho.$$ We can split the multi-variable integral to make calculations easier: $$Q= [\int_{0}^{2\pi} d\phi][\int_{0}^{L} dz][\int_{1}^{\rho_1} (4 \rho^2 \mu C/m^3) d\rho].$$

Evaluate the integrals

Compute the integrals separately: $$\int_{0}^{2\pi} d\phi = 2\pi,$$ $$\int_{0}^{L} dz = L,$$ $$\int_{1}^{\rho_1} (4\rho^2\mu C/m^3) d\rho = 4 \mu C/m^3 \int_{1}^{\rho_1} \rho^2 d\rho.$$ Evaluate the last integral: $$4\mu C/m^3 [(\rho_1^3 / 3) - (1^3 / 3)] = 4\mu C/m^3 (\rho_1^3 - 1)/3.$$

Multiply the results

Now, combine the results of the three integrals: $$Q = 2\pi L\cdot \frac{4\mu C/m^3(\rho_1^3 - 1)}{3}.$$ This gives the total charge in the region of interest. #b) Determine \(D_{\rho}\) at \(\rho=\rho_{1}\)#

Apply Gauss's Law

We can use Gauss's law to determine the electric displacement \(D_\rho\): $$\oint \vec{D} \cdot \mathrm{d}\vec{A} = Q_{enc},$$ where \(Q_{enc}\) is the charge enclosed within the Gaussian surface.

Choose appropriate Gaussian surface

Select a cylindrical Gaussian surface with height \(L\), radius \(\rho_1\), and a base parallel to the \(xy\) plane. Only \(D_\rho\) contributes to the integral, as \(D_z\) is perpendicular to the curved surface area and \(D_\phi\) is perpendicular to the top and bottom surfaces.

Integrating over the surface

The integral becomes: $$\oint \vec{D} \cdot \mathrm{d}\vec{A} = D_\rho \int_{0}^{2\pi} \int_{0}^{L} \rho_1 dL d\phi = 2\pi L \cdot \rho_1 D_\rho.$$ Since we have found the total charge in part (a), we can substitute it into Gauss's Law: $$2\pi L \cdot \rho_1 D_\rho = 2\pi L\cdot \frac{4\mu C/m^3(\rho_1^3 - 1)}{3}.$$

Solve for \(D_\rho\)

Divide both sides by \(2\pi L \cdot \rho_1\): $$D_\rho = \frac{4\mu C/m^3(\rho_1^3 - 1)}{3\rho_1}.$$ This expression provides the value of \(D_\rho\) at \(\rho = \rho_1\). #c) Evaluate \(D_{\rho}\) at specified values of \(\rho\)# Using the result from part (b), we can now evaluate \(D_{\rho}\) for the three values of \(\rho\) given in the problem:

\(\rho = 0.8\) mm

The value of \(\rho\) is less than 1 mm, so no charge is enclosed. Therefore, \(D_\rho = 0\).

\(\rho = 1.6\) mm

Using the obtained formula: $$D_\rho = \frac{4\mu C/m^3(1.6^3 - 1)}{3(1.6)} \approx 1.409\mu C/m^2.$$

\(\rho = 2.4\) mm

Since the volume charge density is zero for \(\rho > 2\) mm, the enclosed charge will be the same as when \(\rho = 2\) mm. Using the formula: $$D_\rho = \frac{4\mu C/m^3(2^3 - 1)}{3(2.4)} \approx 1.667\mu C/m^2.$$