The electric potential in a region that is within 2.00 m of the origin of a rectangular coordinate system is given by V = Axl + Bym + Czn + D, where A, B, C, D, l, m, and n are constants. The units of A, B, C, and D are such that if x, y, and z are in meters, then V is in volts. You measure V and each component of the electric field at four points and obtain these results:

(a) Use the data in the table to calculate A, B, C, D, l, m, and n. (b) What are V and the magnitude of E at the points (0, 0, 0), (0.50 m, 0.50 m, 0.50 m), and (1.00 m, 1.00 m, 1.00 m)?

The electric field E can be obtained if the potential V is known. This is because the electric field is the negative of the gradient of the potential, this is

$$\begin{gathered} \overrightarrow{\mathrm{E}}⃗(\mathrm{x},\mathrm{y},\mathrm{z})=-\nabla \mathrm{V}(\mathrm{x},\mathrm{y},\mathrm{z}) \\ \mathrm{V}(\mathrm{x},\mathrm{y},\mathrm{z})=\mathrm{A}({\mathrm{x}}^2-3{\mathrm{y}}^2+{\mathrm{z}}^2) \end{gathered}$$

localid="1664869151158" $$\begin{gathered} E=-\nabla V \\ =E_x{e}_x+E_y{e}_y+E_z{e}_z \\ =-\left(\frac{\partial V}{\partial x}{e}_x\frac{\partial V}{\partial y}{e}_y\frac{\partial V}{\partial y}{e}_y\right) \\ {E}_x=-\frac{\partial V}{\partial x} \\ {E}_y=-\frac{\partial V}{\partial y} \\ {E}_z=-\frac{\partial V}{\partial Z} \end{gathered}$$

Solve for the first point:

$\begin{array}{l}{V}_1=A{x}^I+B{y}^m+C{Z}^n+D\\ 10=0+0+0+D\end{array}$

So, the constant D is equal to 10 V.

Solve for the second point

$\begin{array}{l}{\mathrm{V}}_2={\mathrm{Ax}}^{\mathrm{t}}+{\mathrm{By}}^{\mathrm{m}}++{\mathrm{Cz}}^{\mathrm{n}}+\mathrm{D}\\ 4=\mathrm{A}\times \left(1\mathrm{m}\right)+0+0+10\\ \mathrm{A}=-6\mathrm{V}.\mathrm{m}\end{array}$

So, the constant A is equal to

Solve for the third point

$\begin{array}{l}{\mathrm{V}}_3={\mathrm{A}}^{\mathrm{t}}+{\mathrm{By}}^{\mathrm{n}}++{\mathrm{C}}^{\mathrm{n}}+\mathrm{D}\\ 6=0+\mathrm{B}\times {\left(1\mathrm{m}\right)}^{\mathrm{m}}+0+10\\ \mathrm{B}=-4\mathrm{V}.{\mathrm{m}}^{-\mathrm{m}}\end{array}$

So, the constant B is equal to

Solve for the forth point

localid="1664867321116" $\begin{array}{l}{\mathrm{V}}_3={\mathrm{A}}^{\mathrm{t}}+{\mathrm{By}}^{\mathrm{n}}++{\mathrm{C}}^{\mathrm{n}}+\mathrm{D}\\ 6=0+\mathrm{B}\times {\left(1\mathrm{m}\right)}^{\mathrm{m}}+0+10\\ C=-2\mathrm{V}.{\mathrm{m}}^{-\mathrm{m}}\end{array}$

So, the constant C is equal to

Solve for l, m and n:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{x}}=-\frac{\partial \mathrm{V}}{\partial \mathrm{x}} \\ 12\mathrm{V}/\mathrm{m}=-6\mathrm{V}.{\mathrm{m}}^{-1}(-/{\left(1\mathrm{m}\right)}^{/-1}) \\ \mathrm{I}=2 \\ {\mathrm{E}}_{\mathrm{x}}=-\frac{\partial \mathrm{V}}{\partial \mathrm{Y}} \\ 12\mathrm{V}/\mathrm{m}=-4\mathrm{V}.{\mathrm{m}}^{-1}(-\mathrm{m}{\left(1\mathrm{m}\right)}^{\mathrm{m}-1}) \\ \mathrm{m}=3 \\ {\mathrm{E}}_{\mathrm{x}}=-\frac{\partial \mathrm{V}}{\partial \mathrm{Z}} \\ 12\mathrm{V}/\mathrm{m}=-2\mathrm{V}.{\mathrm{m}}^{-1}(-\mathrm{n}{\left(1\mathrm{m}\right)}^{\mathrm{n}-1}) \\ \mathrm{n}=6 \end{gathered}$$

The value of constants are A=-6 V.m-2, B=-4 V.m-3, C=-2 V.m-6, D=10 V, l=2, m=3 and n=6.

The equation of V becomes:

$\mathrm{V}=(-6\mathrm{V}\cdot {\mathrm{m}}^{-2}){\mathrm{x}}^2+(-4\mathrm{V}\cdot {\mathrm{m}}^{-3}){\mathrm{y}}^3+(-2\mathrm{V}\cdot {\mathrm{m}}^{-6}){\mathrm{z}}^6+10\mathrm{V}$

Solving for points (0,0,0)

$$\begin{gathered} \mathrm{V}=(-6\mathrm{V}\cdot {\mathrm{m}}^{-2}){\mathrm{x}}^2+(-4\mathrm{V}\cdot {\mathrm{m}}^{-3}){\mathrm{y}}^3+(-2\mathrm{V}\cdot {\mathrm{m}}^{-6}{\mathrm{z}}^2+10\mathrm{V} \\ =(-6\mathrm{V}\cdot {\mathrm{m}}^{-2}){\left(0\mathrm{m}\right)}^2+(-4\mathrm{V}\cdot {\mathrm{m}}^{-3}){\left(0\mathrm{m}\right)}^3+(-2\mathrm{V}\cdot {\mathrm{m}}^{-6}{\left(0\mathrm{m}\right)}^6+10\mathrm{V} \\ =10\mathrm{V} \end{gathered}$$

Ex=0, Ey=0, Ez=0 ………….. [Given]

$\begin{array}{l}\mathrm{E}=\sqrt{{\mathrm{E}}_{\mathrm{x}}^2+{\mathrm{E}}_{\mathrm{y}}^2+{\mathrm{E}}_{\mathrm{z}}^2}\\ =\sqrt{{(0\mathrm{V}.\mathrm{m})}^2+{(0\mathrm{V}.\mathrm{m})}^2+{(0\mathrm{V}.\mathrm{m})}^2}\\ =0\mathrm{V}/\mathrm{m}\end{array}$

The magnitude of V is 10V and the magnitude of E is 0 V/m at the point (0,0,0)

Solving for points (0.5,0.5,0.5)

$$\begin{gathered} \mathrm{V}=(-6\mathrm{V}\cdot {\mathrm{m}}^{-2}){\mathrm{x}}^2+(-4\mathrm{V}\cdot {\mathrm{m}}^{-3}){\mathrm{y}}^3+(-2\mathrm{V}\cdot {\mathrm{m}}^{-6}{\mathrm{z}}^2+10\mathrm{V} \\ =(-6\mathrm{V}\cdot {\mathrm{m}}^{-2}){(0.5\mathrm{m})}^2+(-4\mathrm{V}\cdot {\mathrm{m}}^{-3}){(0.5\mathrm{m})}^3+(-2\mathrm{V}\cdot {\mathrm{m}}^{-6}{(0.5\mathrm{m})}^6+10\mathrm{V} \\ =7.968\mathrm{V} \end{gathered}$$$$\begin{gathered} {\mathrm{E}}_{\mathrm{x}}=-\frac{\partial \mathrm{V}}{\partial \mathrm{x}} \\ =-6\mathrm{V}.{\mathrm{m}}^{-1}(-2\left(0.5\mathrm{m}\right)) \\ 6\mathrm{V}.\mathrm{m} \\ {\mathrm{E}}_{\mathrm{y}}=-\frac{\partial \mathrm{V}}{\partial \mathrm{y}} \\ =-4\mathrm{V}.{\mathrm{m}}^{-1}(-3{\left(0.5\mathrm{m}\right)}^2) \\ 3\mathrm{V}/\mathrm{m} \\ {\mathrm{E}}_{\mathrm{z}}=-\frac{\partial \mathrm{V}}{\partial \mathrm{y}} \\ =-2\mathrm{V}.{\mathrm{m}}^{-6}(-6{\left(0.5\mathrm{m}\right)}^5) \\ 0.375\mathrm{V}/\mathrm{m} \end{gathered}$$

$\begin{array}{l}\mathrm{E}=\sqrt{{\mathrm{E}}_{\mathrm{x}}^2+{\mathrm{E}}_{\mathrm{y}}^2+{\mathrm{E}}_{\mathrm{z}}^2}\\ =\sqrt{{(6\mathrm{V}.\mathrm{m})}^2+{(3\mathrm{V}.\mathrm{m})}^2+{(0.375\mathrm{V}.\mathrm{m})}^2}\\ =6.719\mathrm{V}/\mathrm{m}\end{array}$

The magnitude of V is 7.968V and the magnitude of E is 6.719 V/m at the point (0.5,0.5,0.5)

Solving for points (1,1,1)

$$\begin{gathered} \mathrm{V}=(-6\mathrm{V}\cdot {\mathrm{m}}^{-2}){\mathrm{x}}^2+(-4\mathrm{V}\cdot {\mathrm{m}}^{-3}){\mathrm{y}}^3+(-2\mathrm{V}\cdot {\mathrm{m}}^{-6}){\mathrm{z}}^2+10\mathrm{V} \\ =(-6\mathrm{V}\cdot {\mathrm{m}}^{-2}){\left(1\mathrm{m}\right)}^2+(-4\mathrm{V}\cdot {\mathrm{m}}^{-3}){\left(1\mathrm{m}\right)}^3+(-2\mathrm{V}\cdot {\mathrm{m}}^{-6}{\left(1\mathrm{m}\right)}^6+10\mathrm{V} \\ =-2\mathrm{V} \end{gathered}$$

$$\begin{gathered} {\mathrm{E}}_{\mathrm{x}}=-\frac{\partial \mathrm{V}}{\partial \mathrm{x}} \\ =-6\mathrm{V}.{\mathrm{m}}^{-1}(-2\left(1\mathrm{m}\right)) \\ 6\mathrm{V}.\mathrm{m} \\ {\mathrm{E}}_{\mathrm{y}}=-\frac{\partial \mathrm{V}}{\partial \mathrm{y}} \\ =-4\mathrm{V}.{\mathrm{m}}^{-1}(-3{\left(1\mathrm{m}\right)}^2) \\ 3\mathrm{V}/\mathrm{m} \\ {\mathrm{E}}_{\mathrm{z}}=-\frac{\partial \mathrm{V}}{\partial \mathrm{z}} \\ =-2\mathrm{V}.{\mathrm{m}}^{-6}(-6{\left(1\mathrm{m}\right)}^5) \\ 0.375\mathrm{V}/\mathrm{m} \end{gathered}$$

$\begin{array}{l}\mathrm{E}=\sqrt{{\mathrm{E}}_{\mathrm{x}}^2+{\mathrm{E}}_{\mathrm{y}}^2+{\mathrm{E}}_{\mathrm{z}}^2}\\ =\sqrt{{(12\mathrm{V}.\mathrm{m})}^2+{(12\mathrm{V}.\mathrm{m})}^2+{(12\mathrm{V}.\mathrm{m})}^2}\\ =20.785\mathrm{V}/\mathrm{m}\end{array}$

The magnitude of V is -2 V and the magnitude of E is 20.785 V/m at the point (1,1,1)