Electric Field As Derivative of Potential Video Lessons

Concept

# Problem: The electric potential in a region of space is V = (150x2 – 200y2) V, where x and y are in meters. What are the strength and direction of the electric field at (x, y) = (2.0 m,2.0 m)? Give the direction as an angle cw or ccw (specify which) from the positive x-axis.

###### FREE Expert Solution

We're told that the electric potential in a region of space is:

$\mathbit{V}\mathbf{=}\mathbf{\left(}\mathbf{150}{\mathbit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{200}{\mathbit{y}}^{\mathbf{2}}\mathbf{\right)}\mathbit{V}$

We take the components of the electric field in the x and y directions to be Ex and Ey, respectively.

Therefore,

$\overline{){{\mathbf{E}}}_{{\mathbf{x}}}{\mathbf{=}}{\mathbf{-}}\frac{\mathbf{d}\mathbf{V}}{\mathbf{d}\mathbf{x}}}$

Similarly,

$\overline{){{\mathbf{E}}}_{{\mathbf{y}}}{\mathbf{=}}{\mathbf{-}}\frac{\mathbf{d}\mathbf{V}}{\mathbf{d}\mathbf{y}}}$

The negative sign in these equations is for the potential difference coefficient.

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###### Problem Details

The electric potential in a region of space is V = (150x2 – 200y2) V, where x and y are in meters. What are the strength and direction of the electric field at (x, y) = (2.0 m,2.0 m)? Give the direction as an angle cw or ccw (specify which) from the positive x-axis.