Question

Question: The electric potential at points in an x-yplane is given by$\mathit{v}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{0}\mathit{V}\mathbf{/}{\mathit{m}}^{\mathbf{2}}\mathbf{)}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{V}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}\mathbf{)}{\mathit{y}}^{\mathbf{2}}\mathbf{}$. In unit-vector notation, what is the electric field at the point (role="math" localid="1662092062999" $\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\text{m}}\mathbf{,}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{\text{m}}$)?

Short answer

Answer:

The electric field at the point (3m,2m) is $\left(-12.0\text{V/m}\right)\hat{i}+\left(12.0\text{V/m}\right)\hat{j}$.

Step-by-step solution

The given data

1. The electric potential in an x-y plane $v=\left(2.0{\text{V/m}}^{\text{2}}\right)x^2-\left(3.0{\text{V/m}}^{\text{2}}\right)y^2$,
2. The coordinates of the point $\text{(x,y)}=\text{(3m,2m)}$,

Understanding the concept of the electric field

Using the given data of potential and the coordinate value, we can get the value of the electric field by substituting the given values.

Formula:

The electric field due to the change in potential difference $\overrightarrow{E}=-\frac{\partial V}{\partial s}$, (i)

Calculation of the electric field

Using the given data in equation (i), we can get the value of the x-component of the electric field as follows:

$$\begin{gathered} E_x=\frac{-\partial }{\partial x}\left[\left(2.0{\text{V/m}}^{\text{2}}\right)x^2-\left(3.0{\text{V/m}}^{\text{2}}\right)y^2\right] \\ =-\left(2.0{\text{V/m}}^{\text{2}}\right)\left(2x\right) \end{gathered}$$

Using the given data in equation (i), we can get the value of the y-component of the electric field as follows:

$$\begin{gathered} E_y=\frac{-\partial }{\partial y}\left[\left(2.0{\text{V/m}}^{\text{2}}\right)x^2-\left(3.0{\text{V/m}}^{\text{2}}\right)y^2\right] \\ =\left(3.0{\text{V/m}}^{\text{2}}\right)\left(2y\right) \end{gathered}$$

The net electric field is expressed using above values for the given coordinates as follows:

$$\begin{gathered} \overrightarrow{E}=E_x\hat{i}+E_y\hat{j} \\ =-\left(2.0{\text{V/m}}^{\text{2}}\right)\left(2x\right)\hat{i}+\left(3.0{\text{V/m}}^{\text{2}}\right)\left(2y\right)\hat{j} \\ =-\left(2.0{\text{V/m}}^{\text{2}}\right)\left[2\left(3.0\text{m}\right)\right]\hat{i}+\left(3.0{\text{V/m}}^{\text{2}}\right)\left[2\left(2.0\text{m}\right)\right]\hat{j} \\ =\left(-12.0\text{V/m}\right)\hat{i}+\left(12.0\text{V/m}\right)\hat{j} \end{gathered}$$

Therefore, the net electric field at the point (3m,2m) in unit vector notation is $\left(-12.0\text{V/m}\right)\hat{i}+\left(12.0\text{V/m}\right)\hat{j}$