One of these is an impossible electrostatic field. Which one?

(a) $\mathit{E}\mathbf{=}\mathit{k}\left[xy\hat{x}+2yz\hat{y}+3xz\hat{z}\right]$

(b) $\mathit{E}\mathbf{=}\mathit{k}\mathbf{[}{\mathbf{y}}^{\mathbf{2}}\hat{\mathbf{x}}\mathbf{+}\mathbf{(}\mathbf{2}\mathbf{yz}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}\mathbf{)}\hat{\mathbf{y}}\mathbf{+}\mathbf{2}\mathbf{yz}\hat{\mathbf{z}}\mathbf{]}$.

Here k is a constant with the appropriate units. For the possible one, find the potential, using the origin as your reference point. Check your answer by computing $\mathbf{\nabla }\mathit{V}$. [Hint: You must select a specific path to integrate along. It doesn't matter what path you choose, since the answer is path-independent, but you simply cannot integrate unless you have a definite path in mind.]

The electrostatics field,

(a)$E=k[\mathrm{xy}\hat{\mathrm{x}}+2\mathrm{yz}\hat{\mathrm{y}}+3\mathrm{xz}\hat{\mathrm{z}}]$

(b)$E=k[{\mathrm{y}}^2\hat{\mathrm{x}}+(2\mathrm{yz}+{\mathrm{z}}^2)\hat{\mathrm{y}}+2\mathrm{yz}\hat{\mathrm{z}}]$

The expression to check the existence of the electrostatics field is given as follows.

$\mathbf{\nabla }\mathbf{\times }\mathit{E}\mathbf{=}\mathbf{0}$

The expression to calculate the electric potential is given as follows.

$\mathit{V}\mathbf{=}\mathbf{\int }\mathit{E}\mathit{d}\mathit{l}$

(a)

Determine the existence of the field:

$$\begin{gathered} \nabla \times E=k\left|\begin{array}{ccc}x& y& z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ xy& 2yz& 3zx\end{array}\right| \\ \nabla \times E=k\left[x\left(0-2y\right)+y\left(0-3z\right)+z\left(0-x\right)\right] \\ \nabla \times E=k\left(-2xy+3yz-zx\right) \\ \nabla \times E\ne 0 \end{gathered}$$

Since the condition for the field existence is not satisfied therefore, field$E=k[\mathrm{xy}\hat{\mathrm{x}}+2\mathrm{yz}\hat{\mathrm{y}}+3\mathrm{xz}\hat{\mathrm{z}}]$ impossible.

(b)

Determine the existence of the field,

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

Since the condition for the field existence is satisfied therefore, field$E=k[{\mathrm{y}}^2\hat{\mathrm{x}}+(2\mathrm{yz}+{\mathrm{z}}^2)\hat{\mathrm{y}}+2\mathrm{yz}\hat{\mathrm{z}}]$possible.

Let assume the points $\left(x_0,y_0,z_0\right)$to find the potential for the electric field.

Let assume the V is the potential.

Calculate the value of Edl.

$Edl=k\left[y^{2}dx+\left(2xy+z^2\right)dy+2yzdz\right]$ …… (1)

For the path I which along the x plane,

$\begin{array}{l}dz=0\\ dy=0\end{array}$

Substitute 0 for dz and dy into equation (1).

$$\begin{gathered} Edl=k\left[y^{2}dx+(2xy+z^2)\left(0\right)+2yz\left(0\right)\right] \\ Edl=k{y}^{2}dx \\ Edl=0 \end{gathered}$$

For the path II which along the x-y plane:

$\begin{array}{l}x=x_0\\ y=0to{y}_0\\ z=0\end{array}$

Then

$\begin{array}{l}dx=0\\ dz=0\end{array}$

Substitute 0 for dz and dx into equation (1).

$$\begin{gathered} Edl=k\left[y^2\left(0\right)+(2xy+z^2)dy+2yz\left(0\right)\right] \\ Edl=k(2xy+z^2)dy \\ Edl=2k{x}_{0}ydy \end{gathered}$$

Integrate both the sides,

$$\begin{gathered} \int Edl={\int }_0^{y_0}2k{x}_{0}ydy \\ \int Edl=2k{x}_0{\int }_0^{y_0}ydy \\ \int Edl=2k{x}_0\left(\frac{y_0^2}{2}\right) \\ \int Edl=k{x}_0{y}_0^2 \end{gathered}$$

For the path III which along the y-z plane,

$\begin{array}{l}x=x_0\\ y=y_0\\ z=0\mathrm{to}{z}_0\end{array}$

Then

Substitute 0 for dy and dx into equation (1).

$$\begin{gathered} Edl=k\left[y^2\left(0\right)+(2xy+z^2)\left(0\right)+2yzdz\right] \\ Edl=2kyzdz \\ Edl=2k{y}_{0}zdz \end{gathered}$$

Integrate both the sides,

$$\begin{gathered} \int Edl={\int }_0^{z_0}2k{y}_{0}zdz \\ \int Edl=2k{y}_0{\int }_0^{z_0}zdz \\ \int Edl=2k{y}_0\left(\frac{z_0^2}{2}\right) \\ \int Edl=k{y}_0{z}_0^2 \end{gathered}$$

The net electric potential is given by,

$\begin{array}{l}V(x_0,y_0,z_0)=\int Edl\\ V\left(x_0,y_0,z_0\right)=-k(x_0{y}_0^2+y_0{z}_0^2)\end{array}$

Check the result. Determination of $-\nabla V$.

$$\begin{gathered} -\nabla V=k\left[\frac{\partial }{\partial x}\left(x{y}^2+y{z}^2\right)\hat{x}+\frac{\partial }{\partial y}\left(x{y}^2+y{z}^2\right)\hat{y}+\frac{\partial }{\partial z}\left(x{y}^2+y{z}^2\right)\hat{z}\right] \\ -\nabla V=k\left(y^2\hat{x}+\left(2xy\right)\hat{z}+2yz\hat{z}\right) \\ -\nabla V=E \end{gathered}$$

Hence, the electric potential is $V(x_0,y_0,z_0)=-k(x_0{v}_0^2+y_0{z}_0^2)$and also verified.