Question

(a) Use Maxwell's first equation, \(\nabla \cdot \mathbf{D}=\rho_{v}\), to describe the variation of the electric field intensity with \(x\) in a region in which no charge density exists and in which a nonhomogeneous dielectric has a permittivity that increases exponentially with \(x\). The field has an \(x\) component only; \((b)\) repeat part \((a)\), but with a radially directed electric field (spherical coordinates), in which again \(\rho_{v}=0\), but in which the permittivity decreases exponentially with \(r\).

Short answer

Answer: The expression for the variation of radially directed electric field intensity with radial distance (r) in a region with a nonhomogeneous dielectric is given by \(E_r(r) = \frac{C_2}{\epsilon(r) r^2}\), where \(C_2\) is a constant.

Step-by-step solution

Write Maxwell's first equation in terms of given variables

In this problem, we are given the equation \(\nabla \cdot \mathbf{D}=\rho_{v}\). Since no charge density exists, we have \(\rho_{v}=0\) and the equation becomes \(\nabla \cdot \mathbf{D} = 0\).

Express \(\mathbf{D}\) in terms of electric field intensity \(E_x\) and permittivity

For a nonhomogeneous dielectric with increasing permittivity, we have \(\mathbf{D} = \epsilon \mathbf{E}\). Since the electric field only has an \(x\) component, we can write: \(\mathbf{D} = \epsilon(x) E_x(x) \hat{x}\)

Calculate divergence of \(\mathbf{D}\)

Now, we need to find the divergence of \(\mathbf{D}\): \(\nabla \cdot \mathbf{D} = \frac{\partial (\epsilon(x) E_x(x))}{\partial x}\)

Set the divergence to zero and solve for \(E_x\)

Using the derived divergence from step 3 and the fact that \(\rho_v=0\), we now have: \(\frac{\partial (\epsilon(x) E_x(x))}{\partial x} = 0\) Integrating both sides with respect to \(x\) yields: \(\epsilon(x) E_x(x) = C_1\) We finally get the expression for the variation of electric field intensity with \(x\): \(E_x(x) = \frac{C_1}{\epsilon(x)}\) (b)

Repeat steps 1 and 2 for spherical coordinates

Using Maxwell's first equation in spherical coordinates, the divergence of \(\mathbf{D}\) for a radially varying electric field is given by: \(\nabla \cdot \mathbf{D} = \frac{1}{r^2} \frac{\partial (\epsilon(r) E_r(r) r^2)}{\partial r}\) Since \(\rho_{v} = 0\), the equation becomes: \(\frac{1}{r^2} \frac{\partial (\epsilon(r) E_r(r) r^2)}{\partial r} = 0\)

Solve the equation for \(E_r\)

Now, we need to solve this equation for \(E_r(r)\): \(\frac{\partial (\epsilon(r) E_r(r) r^2)}{\partial r} = 0\) Integrating both sides with respect to \(r\) yields: \(\epsilon(r) E_r(r) r^2 = C_2\) Finally, we get the expression for the variation of radially directed electric field intensity with \(r\): \(E_r(r) = \frac{C_2}{\epsilon(r) r^2}\)