Q53P - Introduction to Electrodynamics

Question

(a) Which of the vectors in Problem 1.15 can be expressed as the gradient of a scalar? Find a scalar function that does the job.

(b) Which can be expressed as the curl of a vector? Find such a vector.

Step-by-Step Solution

Answer

Step 1: Describe the given information

Write the given vectors.

$$\begin{gathered} v_a=x^{2}i+3x{z}^{2}j+(-2xz)k \\ {v}_b=y^{2}i+2yzj+3zxk \\ {v}_c=y^{2}i+(2xy+z^2)j+2yzk \end{gathered}$$

Step 2: Define the line integral

The gradient of a scalar function $\mathit{F}$  is defined as$\mathbf{\nabla }\mathit{F}$   and curl of a vector function $v$  is defined as $\mathbf{\nabla }\mathbf{\times }\mathit{v}$.

Step3: Find the scalar function for part (a)

$$\begin{gathered} \nabla .v_a=\left[\begin{array}{c}\frac{\partial }{\partial x}(x^{2}i+3x{z}^{2}j+(-2xz\left)k\right)+\frac{\partial }{\partial y}(x^{2}i+3x{z}^{2}j+(-2xz\left)k\right)\\ +\frac{\partial }{\partial z}(x^{2}i+3x{z}^{2}j+(-2xz\left)k\right)\end{array}\right] \\ =2x+0-2x \\ =0 \end{gathered}$$

The dot product of vector $v_b$ is obtained as follows:

$$\begin{gathered} \nabla .v_a=\left[\begin{array}{c}\frac{\partial }{\partial x}(y^{2}i+2yzj+3zx k)+\frac{\partial }{\partial y}(y^{2}i+2yzj+3zx k)\\ +\frac{\partial }{\partial z}(y^{2}i+2yzj+3zx k)\end{array}\right] \\ =y+2z+3x \end{gathered}$$

The dot product of vector  $v_c$ is obtained as follows:

$$\begin{gathered} \nabla .v_a=\left[\begin{array}{c}\frac{\partial }{\partial x}(y^{2}i+(2xy+z^2)j+2yz k)+\frac{\partial }{\partial y}(y^{2}i+(2xy+z^2)j+2yz k)\\ +\frac{\partial }{\partial z}(y^{2}i+(2xy+z^2)j+2yz k)\end{array}\right] \\ =0+(2x+0)+2y \\ =2(x+y) \end{gathered}$$

Step: 4 Find the cross product of the given vectors

The cross product of vector $v_a$ is obtained as follows:

$$\begin{gathered} \nabla \times v_a=\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ x^2& 3x{z}^2& -2xz\end{array}\right| \\ =\left[\begin{array}{c}i\left(\frac{\partial }{\partial y}(-2zx)-\frac{\partial }{\partial z}\left(3x{z}^2\right)\right)-j\left(\frac{\partial }{\partial x}(-2xz)-\frac{\partial }{\partial z}\left(x^2\right)\right)\\ +i\left(\frac{\partial }{\partial x}\left(3x{z}^2\right)-\frac{\partial }{\partial y}\left(x^2\right)\right)\end{array}\right]/ \\ =-6xzi+2zj+3z^{2}k \end{gathered}$$

The cross product of vector $v_b$ is obtained as follows:

$$\begin{gathered} \nabla \times v_a=\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ y^2& 2yz& 3zx\end{array}\right| \\ =\left[\begin{array}{c}i\left(\frac{\partial }{\partial y}\left(3zx\right)-\frac{\partial }{\partial z}\left(2yz\right)\right)-j\left(\frac{\partial }{\partial x}\left(3zx\right)-\frac{\partial }{\partial z}\left(y^2\right)\right)\\ +i\left(\frac{\partial }{\partial x}\left(2yz\right)-\frac{\partial }{\partial y}\left(y^2\right)\right)\end{array}\right] \\ =-2yi-3zj-xk \end{gathered}$$

The cross product of vector$v_c$  is obtained as follows:

$$\begin{gathered} \nabla \times v_a=\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ y^2& 2xy+z^2& x\end{array}\right| \\ =\left[\begin{array}{c}i\left(\frac{\partial }{\partial y}\left(2yz\right)-\frac{\partial }{\partial z}(2xy+z^2)\right)-j\left(\frac{\partial }{\partial x}\left(2yz\right)-\frac{\partial }{\partial z}\left(y^2\right)\right)\\ +i\left(\frac{\partial }{\partial x}(2xy+z^2)-\frac{\partial }{\partial y}\left(y^2\right)\right)\end{array}\right] \\ =0 \end{gathered}$$

Out of the all the given functions, the curl of vector  $v_c$ is 0. So, it can be expressed as gradient of scalar function.

Let us assume $\nabla A-v_c$  . Expand $\nabla A-v_c$  as,

 $\frac{\partial A}{\partial x}i+\frac{\partial A}{\partial y}j+\frac{\partial A}{\partial z}k=y^{2}i+(2xy+z^2)j+2yzk$

 

 On comparing the right and left side of the above equation, we obtain,

 $$\begin{gathered} \frac{\partial A}{\partial x}=y^2 \\ A=y^{2}x+g(y,z) \end{gathered}$$

 

Where, $g(y,z)$  is a function of y , z.

 

Differentiate the above function with respect to y.\

 $\frac{\partial A}{\partial y}=2yx+\frac{\partial g(y,z)}{\partial y}$

 

Where,   $g(y,z)$is a function of y , z.

 

comparing the right and left side of the above equation, as,

 $2xy+z^2=2yx+\frac{\partial g(y,z)}{\partial y}$

 

We obtain the result, $\frac{\partial g(y,z)}{\partial y}=z^2$ , On integrating this equation with respect to y on both sides, we get,

$g(y,z)=z^{2}y+f\left(z\right)$ 

 

Thus, the scalar function   can be written as $A=y^{2}x+z^{2}y+f\left(z\right)$ .

 

Differentiate the above function with respect to z.

 $\frac{\partial A}{\partial z}-2yz+f^{\text{'}}\left(z\right)$

 

Comparing the right and left side of the above equation, as,

 $2yz=2yz+f^{\text{'}}\left(z\right)$

 

Obtain the result, $f^{\text{'}}\left(z\right)=0$ , On integrating this equation with respect to z on both sides, we get,

  $f^{\text{'}}\left(z\right)=C$

 

Here, C  is some constant. Thus the scalar function is obtained as $A=y^{2}x+z^{2}y+C$  .

Step 5: Find the vector function for part (b)

Out of the all the given functions, the dot product of vector  $v_a$ is 0. So, it can be expressed as curl of a vector, as divergence of curl is always 0.

 Let us assume $\nabla \times F=v_a$  . Expand  $\nabla \times F=v_a$ as,

$$\begin{gathered} \nabla \times F=v_a \\ \left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ F_x& F_y& F_z\end{array}\right|-x^{2}i+3z^{2}j+(-2xz)k \\ \left(\frac{\partial F_z}{\partial y}-\frac{\partial F_y}{\partial z}\right)i-\left(\frac{\partial F_z}{\partial x}-\frac{\partial F_x}{\partial z}\right)j+\left(\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}\right)k=x^{2}i+3z^{2}j+(-2xz)k \end{gathered}$$

On comparing the right and left side of the above equation, we obtain,

 $$\begin{gathered} \left(\frac{\partial F_z}{\partial y}-\frac{\partial F_y}{\partial z}\right)=x^2 ......\left(1\right) \\ \left(\frac{\partial F_x}{\partial z}-\frac{\partial F_z}{\partial x}\right)=3x{z}^2 .......\left(2\right) \\ \left(\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}\right)=2xz .........\left(3\right) \end{gathered}$$

To calculate $F_x$ ,$F_y$  , assume that $F_z=0$  . Substitute 0 for  $F_z$ int equation (2).

$$\begin{gathered} \left(\frac{\partial F_z}{\partial x}-\frac{\partial \left(0\right)}{\partial z}\right)=-3x{z}^2 \\ \frac{\partial F_z}{\partial x}=-3x{z}^2 \\ {F}_z=-\frac{3}{2}{x}^2{z}^2+f(x,y) \end{gathered}$$

Substitute 0 for $F_x$  int equation (3).

$$\begin{gathered} \left(\frac{\partial F_y}{\partial x}-\frac{\partial \left(0\right)}{\partial }\right)=-2xz \\ \frac{\partial Fy}{\partial x}=-2xz \\ {F}_y=-x^{2}z+g(y,z) \end{gathered}$$

Substitute $-x^{2}z+g(y,z)$  for $F_{y,}-\frac{3}{2}-x^2{z}^2+f(x,y)$ for $F_z$  into equation (1).

$$\begin{gathered} \left(\frac{\partial }{\partial y}\left(-\frac{3}{2}-x^2{z}^2+f(x,y)\right)-\frac{\partial }{\partial z}(-x^{2}z+g(y,z\left)\right)\right)=x^2 \\ \frac{\partial }{\partial y}f(y,z)+x^2-\frac{\partial }{\partial z}g(y,z)=x^2 \\ \frac{\partial }{\partial y}f(y,z)+x^2-\frac{\partial }{\partial z}g(y,z)=0 \end{gathered}$$ 

 

The resulting expression  $\frac{\partial }{\partial y}f(y,z)-\frac{\partial }{\partial z}g(y,z)=0$ is possible only when  .

 $f(y,z)=g(y,z)=0$

Thus, the vector $F=F_{x}i+F_{y}j+F_{z}k$  can be written as $F=x^{2}zj-\frac{3}{2}{x}^2{z}^{2}k$ .