Electric field given electric potential

In summary, the electric field strength at a point (1.00 m, 3.00 m) can be found by taking the negative gradient of the electric potential, V = (210x^2 - 170y^2) V, where x and y are in meters. The gradient operator used is \nabla = \frac{\partial}{\partial x} \mathbf{\hat{i}} + \frac{\partial}{\partial y} \mathbf{\hat{j}} +\frac{\partial}{\partial z} \mathbf{\hat{k}}.
  • #1
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[SOLVED] electric field given electric potential

Homework Statement



The electric potential in a region of space is V = ( 210 x^{2} - 170 y^{2} ) V, where x and y are in meters.

What is the strength of the electric field at ( 1.00 m, 3.00 m ) ?

Homework Equations



E = -dV/ds


The Attempt at a Solution



1. i know i have to find the derivative of the that given equation but i can't figure out where to start...because there are two different variable given...and i don't know in term of which variable i have to find the derivative
 
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  • #2
The E-field is a vector quantity and is given by the negative gradient of the potential.

[tex] \mathbf{E} = -\nabla V [/tex]

the [itex]\nabla[/itex] operator is:

[tex] \nabla = \frac{\partial}{\partial x} \mathbf{\hat{i}} + \frac{\partial}{\partial y} \mathbf{\hat{j}} +\frac{\partial}{\partial z} \mathbf{\hat{k}}[/tex]
 
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  • #3
thank you
 
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