Let a_r, a_ϕ, and a_z be unit vectors along r, ϕ and z directions, respectively in the cylindrical coordinate system. For the electric flux density given by D = (a_r 15 + a_ϕ 2r - a_z 3rz) Coulomb/m2, the total electric flux, in Coulomb, emanating from the volume enclosed by a solid cylinder of radius 3 m and height 5 m oriented along the z-axis with its base at the origin is:

The correct option is 180π.

Step-by-step Explanation:

According to Gauss's Law for electric fields, the total electric flux emanated from a closed surface enclosing a volume V is given by the surface integral of the electric flux density vector D, or equivalently, by the volume integral of the divergence of D (Divergence Theorem):

$\Psi =\underset{S}{∯}\mathbf{D}·d\mathbf{S}=\underset{V}{∰}\left(\nabla ·\mathbf{D}\right)dV$

Step 1: Calculate the divergence of D in cylindrical coordinates

The electric flux density vector is given as:

$\mathbf{D}=15{\mathbf{a}}_r+2r{\mathbf{a}}_{\varphi }-3rz{\mathbf{a}}_z$

where its components are $D_r=15$, $D_{\varphi }=2r$, and $D_z=-3rz$.

The divergence of a vector field in cylindrical coordinates is expressed as:

$\nabla ·\mathbf{D}=\frac{1}{r}\frac{\partial }{\partial r}\left(r{D}_r\right)+\frac{1}{r}\frac{\partial D_{\varphi }}{\partial \varphi }+\frac{\partial D_z}{\partial z}$

Now, let us compute each term individually:

1. First term:

$\frac{1}{r}\frac{\partial }{\partial r}\left(r·15\right)=\frac{1}{r}\left(15\right)=\frac{15}{r}$

2. Second term:

$\frac{1}{r}\frac{\partial }{\partial \varphi }\left(2r\right)=0$

3. Third term:

$\frac{\partial }{\partial z}\left(-3rz\right)=-3r$

Summing these terms gives the divergence:

$\nabla ·\mathbf{D}=\frac{15}{r}-3r$

Step 2: Evaluate the volume integral

The differential volume element in cylindrical coordinates is $dV=rdrd\varphi dz$.

The limits of integration for the solid cylinder of radius $R=3\text{m}$ and height $H=5\text{m}$ based at the origin are:

$r:0\to 3$
$\varphi :0\to 2\pi$
$z:0\to 5$

Thus, the total electric flux is:

$\Psi ={\int }_{z=0}^5{\int }_{\varphi =0}^{2\pi }{\int }_{r=0}^3\left(\frac{15}{r}-3r\right)rdrd\varphi dz$

Simplify the integrand:

$\left(\frac{15}{r}-3r\right)r=15-3r^2$

Separate the integrals:

$\Psi =\left({\int }_0^{5}dz\right)\left({\int }_0^{2\pi }d\varphi \right)\left({\int }_0^3\left(15-3r^2\right)dr\right)$

Now calculate each factor:

1. ${\int }_0^{5}dz=5$

2. ${\int }_0^{2\pi }d\varphi =2\pi$

3. ${\int }_0^3\left(15-3r^2\right)dr={\left(15[r[-[r^3\right)}_0^3=\left(15·3-3^3\right)=45-27=18$

Multiplying these results together:

$\Psi =5\times 2\pi \times 18=180\pi \text{Coulomb}$

Therefore, the total electric flux emanating from the cylinder is 180π Coulombs.