Question

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.)

Answer 1

Answer:

$E = -40\hat{i} + 12\hat{j}$

Explanation:

Given that:

               $V=(4x^{2} - 2y^{2} ) V/m^{2}$ --- (1)

To find:

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

Electric field in plane s is related to V by:

                            $E_{s} =- \frac{\partial V}{\partial {S}}$

For xy plane:

$E_{x} - \frac{\partial V}{\partial x} \\E_{y} - \frac{\partial V}{\partial y}$

Using (1) in above two equations

$E_{x} = -\frac{\partial }{\partial x} (4x^{2} - 2y^{2})\\E_{x} = -8x\\E_{y} = -\frac{\partial }{\partial y} (4x^{2} - 2y^{2})\\E_{y} = 4y$

In vector form

$E = E_{x}\hat{i} + E_{y}\hat{j} \\E = -8x\hat{i} + 4y\hat{j}$

at (5,3)

$E = -40\hat{i} + 12\hat{j}$