In a certain medium, the electric potential is given by $$ V(x)=\frac{\rho_{0}}{a \epsilon_{0}}\left(1-e^{-a x}\right) $$ where \(\rho_{0}\) and \(a\) are constants. ( \(a\) ) Find the electric field intensity, E. \((b)\) Find the potential difference between the points \(x=d\) and \(x=0 .(c)\) If the medium permittivity is given by \(\epsilon(x)=\epsilon_{0} e^{a x}\), find the electric flux density, \(\mathbf{D}\), and the volume charge density, \(\rho_{v}\), in the region. ( \(d\) ) Find the stored energy in the region \((0

Since the electric potential is given by the function V(x), we can find the electric field intensity using the negative of the derivative of V(x) with respect to x: $$ \mathbf{E}(x)=-\frac{dV}{dx} $$ Now, differentiate the given V(x) function: $$ \mathbf{E}(x)=-\frac{d}{dx}\left(\frac{\rho_{0}}{a \epsilon_{0}}\left(1-e^{-a x}\right)\right) $$ Find the derivative and simplify: $$ \mathbf{E}(x)=-\frac{\rho_{0}}{a \epsilon_{0}}\left(-a e^{-a x}\right) = \frac{\rho_{0}}{\epsilon_{0}}e^{-a x} $$

To find the potential difference between two points, we can subtract the potential at the first point from the potential at the second point: $$ V(x=d)-V(x=0)=\left(\frac{\rho_{0}}{a \epsilon_{0}}\left(1-e^{-a d}\right)\right) - \left(\frac{\rho_{0}}{a \epsilon_{0}}\left(1-e^{0}\right)\right) $$ Now simplify the expression to get: $$ V(d)-V(0)=\frac{\rho_{0}}{a \epsilon_{0}}\left(1-e^{-a d}-1+1\right)=\frac{\rho_{0}(1-e^{-ad})}{a \epsilon_{0}} $$

The electric flux density can be found using Gauss's law in differential form: $$ \mathbf{D}(x)=\epsilon(x)\mathbf{E}(x) $$ Using the given values of ε(x) and E(x) from the problem and substituting them into the equation, we have: $$ \mathbf{D}(x)=\epsilon_{0}e^{ax}\left(\frac{\rho_{0}}{\epsilon_{0}}e^{-a x}\right)=\rho_0 $$ As the electric flux density D is constant, the volume charge density ρv can also be found using the relationship between D and ρv: $$ \nabla \cdot \mathbf{D} = \rho_v $$ Here, since D is constant and does not depend on position x, the divergence of D is 0. Therefore, the volume charge density ρv is: $$ \rho_v = 0 $$

The stored energy density in a dielectric medium is given by: $$ u=\frac{1}{2}\mathbf{D}(x) \cdot \mathbf{E}(x) $$ Substitute the values of D(x) and E(x) in the equation from previous parts: $$ u=\frac{1}{2}\rho_0 \left(\frac{\rho_{0}}{\epsilon_{0}}e^{-a x}\right)=\frac{\rho_0^2}{2\epsilon_0}e^{-a x} $$ To find the stored energy in the region, integrate the energy density over the volume of the region: $$ W_{stored}=\int_{0}^{d}\int_{0}^{1}\int_{0}^{1}u \, dy \, dz \, dx $$ Perform the integration with respect to y and z first: $$ W_{stored}=\int_{0}^{d}\left(\frac{\rho_0^2}{2\epsilon_0}e^{-a x}\right) dx $$ Now integrate with respect to x and solve: $$ W_{stored}=-\frac{\rho_0^2}{2a\epsilon_0}\left[e^{-ad}-1\right] $$