Equipotential Surfaces and Electric Field

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The discussion revolves around calculating the electric field from given equipotential surfaces. The magnitude of the electric field was determined to be 559 V/M using the formula E = -V/R. The main challenge was finding the correct direction of the electric field, which initially caused confusion but was eventually resolved. The participant also expressed confidence in solving the distance change in potential question. Overall, the focus was on understanding the relationship between electric fields and equipotential surfaces.
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Homework Statement



A given system has the equipotential surfaces shown in the figure .
http://session.masteringphysics.com/problemAsset/1122530/1/Walker.20.39.jpg

A)What is the magnitude of the electric field?
B)What is the direction of the electric field? (in degrees from the +x axis)
C)What is the shortest distance one can move to undergo a change in potential of 1.50 V?


Homework Equations


E=-V/R


The Attempt at a Solution


I found the answer to the Part A:
E= -V/R, R=4sin(26.565)
E=559 V/M

I cannot wrap my head around the answer to Part B, though. I thought it would simply be 180-25.65, but that's wrong. Any help?

I also know how to do Part C, so just Part B would be great!
 
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Answers that are also not correct are:
153
63.4
117.
 
Nevermind, I was doing the angle perpendicular in the wrong direction. I got the answer.
 

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