The electric potential at x = 3.0 m is 120 V, and the electric potential at x = 5.0 m is 190 V. What is the electric field in this region, assuming it's constant?
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2 answers

Please show some work of your own. EField = (Voltage change)/(separation)
when the separation is measured perpendicular to equipotential surfaces.
In your case, V changes by 70 V in 2.0 meters.
In this case, the field in in the x direction 👍
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exactly what drwls said.
Electric field =
(change in charge)/(change in position)
So in this case in particular, you have:
 (190V120V)/(5.00m3.00m)
=  (70V)/(2.00m)
= 35V/m or 35 N/C
*V/m = N/C* 👍
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