Questions: Questão 2 Um cilindro encontra-se carregado com uma densidade volumétrica de carga igual a (3 r+r^3), dada em coordenadas cilíndricas. Encontre a carga encerrada no volume definido pela região E(r, θ, z) 0 ≤ r ≤ 1 ; 0 ≤ z ≤ 1 ; 0 ≤ θ ≤ 2 π Dado: dV = r dz dr dθ Anexo - Consulte a imagem em melhor resolução no final do cadernos de questöes. Assinale a alternativa correta.

Solution Steps

We need to find the total charge enclosed in a cylindrical volume with a given charge density. The charge density is given as a function of the radial coordinate \( r \) in cylindrical coordinates: \(\rho(r) = 3r + r^3\). The volume of interest is defined by the region \(0 \leq r \leq 1\), \(0 \leq z \leq 1\), and \(0 \leq \theta \leq 2\pi\).

The total charge \( Q \) enclosed in the volume can be found by integrating the charge density over the specified volume. The differential volume element in cylindrical coordinates is given by \( dV = r \, dz \, dr \, d\theta \).

The integral to find the total charge is: \[ Q = \int_0^{2\pi} \int_0^1 \int_0^1 \rho(r) \, r \, dz \, dr \, d\theta \]

Substitute \(\rho(r) = 3r + r^3\) into the integral: \[ Q = \int_0^{2\pi} \int_0^1 \int_0^1 (3r + r^3) \, r \, dz \, dr \, d\theta \]

Simplify the integrand: \[ Q = \int_0^{2\pi} \int_0^1 \int_0^1 (3r^2 + r^4) \, dz \, dr \, d\theta \]

First, integrate with respect to \( z \): \[ Q = \int_0^{2\pi} \int_0^1 (3r^2 + r^4) \, [z]_0^1 \, dr \, d\theta = \int_0^{2\pi} \int_0^1 (3r^2 + r^4) \, dr \, d\theta \]

Next, integrate with respect to \( r \): \[ Q = \int_0^{2\pi} \left[ \frac{3r^3}{3} + \frac{r^5}{5} \right]_0^1 \, d\theta = \int_0^{2\pi} \left( r^3 + \frac{r^5}{5} \right)_0^1 \, d\theta \] \[ = \int_0^{2\pi} \left( 1 + \frac{1}{5} \right) \, d\theta = \int_0^{2\pi} \frac{6}{5} \, d\theta \]

Finally, integrate with respect to \( \theta \): \[ Q = \left[ \frac{6}{5} \theta \right]_0^{2\pi} = \frac{6}{5} \times 2\pi = \frac{12\pi}{5} \]

Final Answer