The electric potential at points in an xy plane is given by V = (4.0 V/m^2)x^2 − (2.0 V/m^2)y^2. In unit-vector notation, what is the electric field at the point (5.0m, 3.0m)? (Express your answer in vector form.)

Respuesta :

Answer:

[tex]E = -40\hat{i} + 12\hat{j}[/tex]

Explanation:

Given that:

               [tex]V=(4x^{2}  - 2y^{2} ) V/m^{2}[/tex] --- (1)

To find:

Electric field at point (5,3) in xy plane.

Electric field in plane s is related to V by:

                            [tex]E_{s} =- \frac{\partial V}{\partial {S}}[/tex]

For xy plane:

[tex]E_{x} - \frac{\partial V}{\partial x} \\E_{y} - \frac{\partial V}{\partial y}[/tex]

Using (1) in above two equations

[tex]E_{x} = -\frac{\partial  }{\partial x} (4x^{2} - 2y^{2})\\E_{x} = -8x\\E_{y} = -\frac{\partial }{\partial y} (4x^{2} - 2y^{2})\\E_{y} = 4y[/tex]

In vector form

[tex]E = E_{x}\hat{i} + E_{y}\hat{j} \\E = -8x\hat{i} + 4y\hat{j}[/tex]

at (5,3)

[tex]E = -40\hat{i} + 12\hat{j}[/tex]