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Chapter 3: Problem 6

In free space, a volume charge of constant density \(\rho_{v}=\rho_{0}\) exists within the region \(-\infty

Short Answer

Expert verified
Answer: The electric field \(\mathbf{E}\) everywhere in space is given by: \(\mathbf{E} = (0, 0, \frac{\rho_0 \cdot w}{\epsilon_0})\)

Step by step solution

01

Choose a Gaussian surface

To apply Gauss's law, we will choose a Gaussian surface that will make the calculations easier. Since the volume is infinite in x and y directions and finite in the z direction, we can choose a cuboid Gaussian surface with a width \(w\) along the z-axis, height \(h\) along the y-axis, and length \(l\) along the x-axis, centered at the origin.
02

Apply Gauss's law for \(\mathbf{D}\)

Gauss's law states that \(\oint_{S} \mathbf{D} \cdot d\mathbf{a} = Q_{free}\), where \(S\) is the Gaussian surface and \(Q_{free}\) is the charge contained within the surface. For our cuboid Gaussian surface, we have six faces. We can see that the x and y components of \(\mathbf{D}\) should be zero, as there is no preferred direction in these dimensions. Thus, we will focus on the z-component, and our Gaussian surface integral becomes: \(\oint_{S} D_z d\mathbf{a} = Q_{free}\) Now we find the total charge enclosed by the Gaussian surface: \(Q_{free} = \rho_0 \cdot V = \rho_0 \cdot (w \cdot h \cdot l)\) where \(V\) is the volume of the Gaussian surface.
03

Calculate the electric displacement vector \(\mathbf{D}\)

We will now calculate the electric displacement \(\mathbf{D}\) through the Gaussian surface. Since \(D_x\) and \(D_y\) are zero, we only need to consider \(D_z\): \(\oint_{S} D_z d\mathbf{a} = D_z (\text{top face} - \text{bottom face})\) The top face has an area of \(l \cdot h\) and is parallel to the z-axis, so its contribution is \(D_z\cdot(l\cdot h)\). Similarly, the bottom face gives \(-D_z\cdot(l\cdot h)\). Plugging these into Gauss's law: \(D_z(l\cdot h) - D_z(l\cdot h) = \rho_0 \cdot (w \cdot h \cdot l)\) We see that \(D_z = \rho_0 \cdot w\). This gives us the electric displacement vector: \(\mathbf{D} = (0, 0, \rho_0 \cdot w)\)
04

Find the electric field \(\mathbf{E}\)

We have the electric displacement vector \(\mathbf{D}\), and we know its relationship with the electric field \(\mathbf{E}\): \(\mathbf{D} = \epsilon_0 \mathbf{E}\), where \(\epsilon_0\) is the vacuum permittivity. Hence, the electric field is given by: \(\mathbf{E} = \frac{\mathbf{D}}{\epsilon_0} = \frac{(0, 0, \rho_0 \cdot w)}{\epsilon_0} = (0, 0, \frac{\rho_0 \cdot w}{\epsilon_0})\) Finally, we have found the electric field \(\mathbf{E}\) everywhere in space: \(\mathbf{E} = (0, 0, \frac{\rho_0 \cdot w}{\epsilon_0})\)

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Most popular questions from this chapter

Let \(\mathbf{D}=5.00 r^{2} \mathbf{a}_{r} \mathrm{mC} / \mathrm{m}^{2}\) for \(r \leq 0.08 \mathrm{~m}\) and \(\mathbf{D}=0.205 \mathrm{a}_{r} / r^{2} \mu \mathrm{C} / \mathrm{m}^{2}\) for \(r \geq 0.08 \mathrm{~m} .(a)\) Find \(\rho_{v}\) for \(r=0.06 \mathrm{~m} .(b)\) Find \(\rho_{v}\) for \(r=0.1 \mathrm{~m} .(c)\) What surface charge density could be located at \(r=0.08 \mathrm{~m}\) to cause \(\mathbf{D}=0\) for \(r>0.08 \mathrm{~m} ?\)

In a region in free space, electric flux density is found to be $$ \mathbf{D}=\left\\{\begin{array}{lr} \rho_{0}(z+2 d) \mathbf{a}_{z} \mathrm{C} / \mathrm{m}^{2} & (-2 d \leq z \leq 0) \\ -\rho_{0}(z-2 d) \mathbf{a}_{z} \mathrm{C} / \mathrm{m}^{2} & (0 \leq z \leq 2 d) \end{array}\right. $$ Everywhere else, \(\mathbf{D}=0 .\left(\right.\) a) Using \(\nabla \cdot \mathbf{D}=\rho_{v}\), find the volume charge density as a function of position everywhere. (b) Determine the electric flux that passes through the surface defined by \(z=0,-a \leq x \leq a,-b \leq y \leq b\). (c) Determine the total charge contained within the region \(-a \leq x \leq a\), \(-b \leq y \leq b,-d \leq z \leq d .(d)\) Determine the total charge contained within the region \(-a \leq x \leq a,-b \leq y \leq b, 0 \leq z \leq 2 d\).

An electric field in free space is \(\mathbf{E}=\left(5 z^{2} / \epsilon_{0}\right) \hat{\mathbf{a}}_{z} \mathrm{~V} / \mathrm{m}\). Find the total charge contained within a cube, centered at the origin, of \(4-\mathrm{m}\) side length, in which all sides are parallel to coordinate axes (and therefore each side intersects an axis at \(\pm 2\) ).

The cylindrical surface \(\rho=8 \mathrm{~cm}\) contains the surface charge density, \(\rho_{S}=\) \(5 e^{-20|z|} \mathrm{nC} / \mathrm{m}^{2} .(a)\) What is the total amount of charge present? \((b)\) How much electric flux leaves the surface \(\rho=8 \mathrm{~cm}, 1 \mathrm{~cm}

(a) A flux density field is given as \(\mathbf{F}_{1}=5 \mathbf{a}_{z} .\) Evaluate the outward flux of \(\mathbf{F}_{1}\) through the hemispherical surface, \(r=a, 0<\theta<\pi / 2,0<\phi<2 \pi\) (b) What simple observation would have saved a lot of work in part \(a ?\) (c) Now suppose the field is given by \(\mathbf{F}_{2}=5 z \mathbf{a}_{z} .\) Using the appropriate surface integrals, evaluate the net outward flux of \(\mathbf{F}_{2}\) through the closed surface consisting of the hemisphere of part \(a\) and its circular base in the \(x y\) plane. ( \(d\) ) Repeat part \(c\) by using the divergence theorem and an appropriate volume integral.
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