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

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

Step-by-step 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) - (\partial V)/(\partial x) \\E_(y) - (\partial V)/(\partial y)`

Using (1) in above two equations

`E_(x) = -(\partial  )/(\partial x) (4x^(2) - 2y^(2))\\E_(x) = -8x\\E_(y) = -(\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}`