Question

In a certain region of space the electric potential is given by$\mathbf{V}\mathbf{=}{\mathbf{Ax}}^{\mathbf{2}}\mathbf{y}\mathbf{-}{\mathbf{Bxy}}^{\mathbf{2}}$, where$\mathbf{A}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{V}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$and$\mathit{B}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{00}\mathit{V}\mathbf{/}{\mathit{m}}^{\mathbf{3}}$. Calculate the magnitude and direction of the electric field at the point in the region that has coordinates x = 2.00 m, y = 0.400 m, and z = 0.

Short answer

$$\begin{gathered} \mathrm{E}=-6.72\mathrm{i}-7.2\mathrm{j}\mathrm{N}/\mathrm{C} \\ \mathrm{E}=9.85\mathrm{N}/\mathrm{C} \\ \mathrm{\theta }={227}^{\circ } \end{gathered}$$

Step-by-step solution

parameters.

$$\begin{gathered} \mathrm{V}={\mathrm{Ax}}^2\mathrm{y}-{\mathrm{Bxy}}^2 \\ \mathrm{A}=5.00\mathrm{V}/{\mathrm{m}}^3 \\ \mathrm{B}=8.00\mathrm{V}/{\mathrm{m}}^3 \\ \mathrm{x}=2.00\mathrm{m} \\ \mathrm{y}=0.4\mathrm{m} \end{gathered}$$

Calculating the magnitude and direction of the electric field.

The electric field is given by

$$\begin{gathered} \mathit{E}\mathbf{=}\mathbf{-}\mathbf{∆}\mathit{v} \\ \mathbf{=}{\mathit{E}}_{\mathbf{x}}\mathbf{\$}\mathbf{+}{\mathit{E}}_{\mathbf{y}}\mathbf{\$} \end{gathered}$$

The electric field is equal to the negative gradient of the potential so

$E_x=-\frac{\partial V}{\partial x}=-2Axy+B{y}^2$

Also in y direction

$E_y=-\frac{\partial V}{\partial y}=-A{x}^2+2Bxy$

The electric field is given by

$E=[-2Axy+B{y}^2]i^+[-A{x}^2+2Bxy]j^$

The electric field E(2.0,0.4,0) is given by

localid="1664280297072" $$\begin{gathered} E\left(2.0,0.4,0\right)=\left[\left(-2\right)\left(5\right)\left(2\right)\left(0.4\right)+\left(8\right){\left(0.4\right)}^2\right]i\$+\left[-\left(5\right){\left(2\right)}^2+2\left(8\right)\left(0.4\right)\left(2\right)\right] \\ =-6.72i\$-7.2j\$NIC \end{gathered}$$

The magnitude E I given by

$E=\sqrt{(7.2{)}^2+(6.72{)}^2)}=9.85N/C$

Since both components are negative the angle lies in the 3rd quadrant so add to magnitude of angle

$\theta =180+\arctan \frac{13.6}{/6.72}={227}^{\circ }$