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