Answer to Question #225317 in Electricity and Magnetism for Anuj

Question

Question 2

(a) Given that A ~ = (x + 2y + 4z)ˆ

i + (2x - 3y- z)ˆj +(4x-y + 2z)kˆ

(i). Show that the vector field A ~ is irrotational



(ii). Find the scalar potential φ such that A~ = ∇φ, if φ(0, 0, 0) = 1

Answer 1

ANSWER

(i) The vector field $\overrightarrow { A }$ is irrotational because$curl\ \overrightarrow { A } =\ \nabla \times \overrightarrow { A } =\left| \begin{matrix} \overrightarrow { i } & \overrightarrow { j } & \overrightarrow { k } \quad \\ \frac { \partial }{ \partial x } & \frac { \partial }{ \partial y } & \frac { \partial }{ \partial z } \\ (x+2y+4z) & \left( 2x-3y-z \right) & \left( 4x-y+2z \right) \end{matrix} \right| =$

$=\left[ \frac { \partial \left( 4x-y+2z \right) }{ \partial y } -\frac { \partial \left( 2x-3y-z \right) }{ \partial z } \right] \overrightarrow { i } +\left[ \frac { \partial (x+2y+4z) }{ \partial z } -\frac { \partial \left( 4x-y+2z \right) }{ \partial x } \right] \overrightarrow { j } +\left[ \frac { \partial \left( 2x-3y-z \right) }{ \partial x } -\frac { \partial (x+2y+4z) }{ \partial y } \right] \overrightarrow { k } = \\ =\left( -1+1 \right) \overrightarrow { i } +\left( 4-4 \right) \overrightarrow { j } +\left( 2-2 \right) \overrightarrow { k } =0\overrightarrow { i } +0\overrightarrow { j } +0\overrightarrow { k } .\\$

(ii) Therefore , there is scalar potential $\varphi$ such that $\overrightarrow { A } =\nabla \varphi$ :

$\frac { \partial \varphi }{ \partial x } =x+2y+4z\Rightarrow \varphi \left( x,y,z \right) =\frac { { x }^{ 2 } }{ 2 } +2xy+4zx+\alpha (y,z)\Rightarrow \frac { \partial \varphi }{ \partial y } =2x+\frac { \partial \alpha (y,z) }{ \partial y } \\$

Since $\frac { \partial \varphi }{ \partial y } =2x-3y-z\\$ , then $2x+\frac { \partial \alpha (y,z) }{ \partial y } =2x-3y-z\\$ or $\alpha (y,z)=-\frac { 3{ y }^{ 2 } }{ 2 } -zy+\ \beta (z)\\$ and $\varphi \left( x,y,z \right) =\frac { { x }^{ 2 } }{ 2 } +2xy+4zx-\frac { 3{ y }^{ 2 } }{ 2 } -zy+\ \beta (z)\\$ .

Thus, $\frac { \partial \varphi }{ \partial z } =4x-y+{ \beta }^{ ' }{ (z) }=4x-y+2z\\$ $\Rightarrow { \beta }^{ ' }{ (z)=2z },\quad \beta (z)=\ { z }^{ 2 }+\gamma \\$ .

Finally, we have $\varphi \left( x,y,z \right) =\frac { { x }^{ 2 } }{ 2 } +2xy+4zx-\frac { 3{ y }^{ 2 } }{ 2 } -zy+{ z }^{ 2 }+\gamma \\$ .

Since $\varphi \left( 0,0,0 \right) =1$ ,then $\gamma =1\ .\\$ So

$\varphi \left( x,y,z \right) =\frac { { x }^{ 2 } }{ 2 } +2xy+4zx-\frac { 3{ y }^{ 2 } }{ 2 } -zy+{ z }^{ 2 }+1.\\$