91影视

To find the electric field E, we need to compute the negative gradient of the potential V. In cylindrical coordinates, the gradient is expressed as: $$ \nabla V = \frac{\partial V}{\partial \rho} \boldsymbol{e_{\rho}} + \frac{1}{\rho}\frac{\partial V}{\partial \phi} \boldsymbol{e_{\phi}} + \frac{\partial V}{\partial z} \boldsymbol{e_{z}} $$ Since the potential function V given depends only on 蟻: $$ \mathbf{E} = -\nabla V = -\frac{\partial V}{\partial \rho} \boldsymbol{e_{\rho}} $$ Now, let's compute this derivative: $$ \frac{\partial V}{\partial \rho} = \frac{\partial (80 \rho^{0.6})}{\partial \rho} = 80 \cdot 0.6 \rho^{-0.4} $$ Plug this result into the expression for the electric field E: $$ \mathbf{E} = - (80 \cdot 0.6 \rho^{-0.4}) \boldsymbol{e_{\rho}} = -48 \rho^{-0.4} \boldsymbol{e_{\rho}} $$

Next, we will use Gauss's Law to find the volume charge density. Gauss's Law states: $$ \oint \mathbf{E} \cdot d\mathbf{a} = \frac{Q_{enc}}{\epsilon_0} $$ Consider an infinitesimal cylindrical surface of radius 蟻, height dz and thickness d蟻. Then, since the electric field has only a 蟻 component: $$ \oint \mathbf{E} \cdot d\mathbf{a} = E_\rho(\rho) (2\pi \rho dz) $$ The volume enclosed by the infinitesimal surface is dV = 蟻 d蟻 dz d蠁. So, the enclosed charge dQ is: $$ dQ = \rho_v \cdot dV = \rho_v \cdot (\rho d蟻 dz d蠁) $$ Applying Gauss's Law, we get: $$ E_\rho(\rho)(2\pi \rho dz) = \frac{\rho_v (\rho d蟻 dz d蠁)}{\epsilon_0} $$ Therefore, 蟻_v can be found as: $$ \rho_v = \frac{\epsilon_0 E_\rho(\rho)}{\rho} $$ Plug E and 蟻 (0.5 m) into the expression for 蟻_v: $$ \rho_v = \frac{(8.854\times10^{-12})(-48 \cdot 0.5^{-0.4})}{0.5} = -8.854 \times 10^{-12} \cdot 96 \times 0.5^{-1.4} \thickspace C/m^3 $$

To calculate the total charge within the closed surface defined by 蟻 = 0.6 m and 0 < z < 1, we integrate the volume charge density 蟻_v over this volume. Note that the charge density 蟻_v depends only on the distance 蟻. Let's set up the integral: $$ Q = \int_{0}^{2\pi} d\phi \int_{0}^{1} dz \int_{0}^{0.6} \rho_v (\rho) \rho d蟻 $$ Plug 蟻_v into the integral and evaluate: $$ Q = -8.854 \times 10^{-12} \cdot 96 \int_{0}^{2\pi} d\phi \int_{0}^{1} dz \int_{0}^{0.6} \rho^{-1.4} \rho d蟻 = -8.854 \times 10^{-12} \cdot 96 \cdot 2\pi \int_{0}^{1} dz \left[\frac{1}{0.4} \rho^{0.6} \right]_0^{0.6} $$ Calculate the result: $$ Q = -8.854 \times 10^{-12} \cdot 96 \cdot 2\pi \int_{0}^{1} dz (\frac{1}{0.4}(0.6^{0.6} - 0)) = -8.854 \times 10^{-12} \cdot 96 \cdot 2\pi \cdot 1 \cdot \frac{1}{0.4} \cdot 0.6^{0.6} \thickspace C $$ To summarize the results: a) Electric field E: $$ \mathbf{E} = -48 \rho^{-0.4} \boldsymbol{e_{\rho}} $$ b) Volume charge density at 蟻 = 0.5 m: $$ \rho_v = -8.854 \times 10^{-12} \cdot 96 \times 0.5^{-1.4} \thickspace C/m^3 $$ c) Total charge within the closed surface: $$ Q = -8.854 \times 10^{-12} \cdot 96 \cdot 2\pi \cdot 1 \cdot \frac{1}{0.4} \cdot 0.6^{0.6} \thickspace C