Electric potential ( with angle)

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SUMMARY

The electric potential in the given region is defined as V = 200 / √(x² + y²). To determine the electric field at the point (2m, 1m), the negative derivative of the potential was calculated, yielding components Ex = 35.78 and Ey = 17.89. The angle of the electric field was found using the arctangent function, resulting in an angle of 26.6 degrees. The discussion clarified that the angle is measured counterclockwise from the positive x-axis, confirming its position in the first quadrant due to both components being positive.

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Homework Statement


The electric potential of a region of space is V= 200 / sq root (x^2 + y^2), where x and y are in meters. What are the strength and direction of the electric field at (x,y) = (2m,1m)?

Homework Equations

The Attempt at a Solution


I took the negative derivative of V to find what the electric field would be. Then i substituted in (2,1)
200(2)(5)^-3/2 x + 200(5)^-3/2
=35.78x + 17.89 y

im having trouble figuring out where the angle should be. i do this a lot.

if i take the arctan or tan-1 of Ey/Ex

i get an angle of 26.6 degrees. Now is that 26.6 degrees ccw above the x-axis??
im having trouble figuring out where the final electric field will end up.
when i looked at the signs of x,y theyre both negative in the problem which means it would be quadrant 3.
even when i do 360-26.6 it ends in quadrant 4 which doesn't make sense

Please i don't need any help with the problem just with some help understanding the angle
 
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nevermind the x and y components are both positive so it falls in quadrant I.

thread closed
 

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