Electric Field As Derivative of Potential Video Lessons

Concept

# Problem: (II) The electric potential in a region of space varies as V=by/(a2+y2). Determine E.

###### FREE Expert Solution

The electric field, E is:

$\overline{){\mathbf{E}}{\mathbf{=}}{\mathbf{-}}{\mathbf{\nabla }}{\mathbf{V}}}$

But,

${\mathbf{\nabla }}{\mathbf{=}}{\mathbit{i}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{x}}{\mathbf{+}}{\mathbit{j}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{y}}{\mathbf{+}}{\mathbit{k}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{z}}$

Now we can express E as:

$\overline{){\mathbf{E}}{\mathbf{=}}{\mathbf{-}}{\mathbf{\left(}}{\mathbf{i}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{x}}{\mathbf{+}}{\mathbf{j}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{y}}{\mathbf{+}}{\mathbf{k}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{z}}{\mathbf{\right)}}{\mathbf{V}}}$, where V is the electric potential in the region of space.

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###### Problem Details

(II) The electric potential in a region of space varies as V=by/(a2+y2). Determine E.