Q15P - Introduction to Electrodynamics

Question

Calculate the divergence of the following vector functions:

$$\begin{gathered} \mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{ }{\mathit{v}}_{\mathbf{a}}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\hat{\mathbf{x}}\mathbf{+}\mathbf{3}\mathit{x}{\mathit{z}}^{\mathbf{2}}\hat{\mathbf{y}}\mathbf{ }\mathbf{-}\mathbf{2}\mathit{x}\mathit{z}\hat{\mathbf{z}} \\ \mathbf{\left(}\mathit{b}\mathbf{\right)}\mathbf{ }{\mathit{v}}_{\mathbf{b}}\mathbf{=}\mathit{x}\mathit{y}\hat{\mathbf{x}}\mathbf{+}\mathbf{2}\mathit{y}\mathit{z}\hat{\mathbf{y}}\mathbf{+}\mathbf{3}\mathit{z}\mathit{x}\hat{\mathbf{z}} \\ \mathbf{\left(}\mathit{c}\mathbf{\right)}\mathbf{ }{\mathit{v}}_{\mathbf{c}}\mathbf{=}{\mathit{y}}^{\mathbf{2}}\hat{\mathbf{x}}\mathbf{+}\mathbf{(}\mathbf{2}\mathit{x}\mathit{y}\mathbf{+}{\mathit{z}}^{\mathbf{2}}\mathbf{)}\mathbf{+}\mathbf{2}\mathit{y}\mathit{z}\hat{\mathbf{z}} \end{gathered}$$

Step-by-Step Solution

Answer

(a) The divergence is 0.

(b) The divergence is ${}^{3\mathrm{x}+\mathrm{y}+2\mathrm{z}}$.

(c) The divergence is${}^{2( \mathrm{x}+\mathrm{y} )}$.

Step 1: Write the expression for the divergence of the function.

Determine the divergence of the function.

$$\begin{gathered} \nabla .v=\left(\hat{x}\frac{\partial }{\partial x}+\hat{y}\frac{\partial }{\partial y}+\hat{z}\frac{\partial }{\partial z}\right)(v_x\hat{x}+v_y\hat{y}+v_z\hat{z}) \\ \mathbf{ }\mathbf{=}\frac{\mathbf{\partial }{\mathbf{v}}_{\mathbf{x}}}{\mathbf{\partial }\mathbf{x}}\mathbf{+}\hat{\mathbf{y}}\frac{\mathbf{\partial }{\mathbf{v}}_{\mathbf{y}}}{\mathbf{\partial }\mathbf{y}}\mathbf{+}\hat{\mathbf{z}}\frac{\mathbf{\partial }{\mathbf{v}}_{\mathbf{z}}}{\mathbf{\partial }\mathbf{z}} \end{gathered}$$

Step 2: Solve for the divergence of part (a).

Consider the given vector is,

$v_a=x^2 \hat{x}+3x{z}^{2}y-2xz\hat{z}$

Solve for the divergence.

$$\begin{gathered} \nabla .v_a=\frac{\partial }{\partial x}{x}^2+\frac{\partial }{\partial y}3x{z}^2-\frac{\partial }{\partial z}2xz \\ =2x+0-2x \\ =0 \end{gathered}$$

Therefore, the divergence is 0.

Step 3: Solve for the divergence of part (b).

Consider the given vector is,

$v_b=xy\hat{x}+2y{z}^{2}y+3zx\hat{z}$

Solve for the divergence.

$$\begin{gathered} \nabla .v_a=\frac{\partial }{\partial x}xy+\frac{\partial }{\partial y}2yz-\frac{\partial }{\partial z}xz \\ =3x+y+2z \end{gathered}$$

Therefore, the divergence is ${}^{3\mathrm{x}+\mathrm{y}+2\mathrm{z}}$

Step 4: Solve for the divergence of part (c).

Consider the given vector is,

$v_a=y^2 \hat{x}+(2xy+z^2)+2yz\hat{z}$

Solve for the divergence.

$$\begin{gathered} \nabla .v_a=\frac{\partial }{\partial x}{y}^2+\frac{\partial }{\partial y}(2xy+z^2)+\frac{\partial }{\partial z}2yz \\ =0+(2x+0)+2y \\ =2(x+y) \end{gathered}$$

Therefore, the divergence is${}^{2(\mathrm{x}+\mathrm{y})}$.