# physics

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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?

• physics -

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

• physics -

exactly what drwls said.

Electric field =
-(change in charge)/(change in position)

So in this case in particular, you have:
- (190V-120V)/(5.00m-3.00m)
= - (70V)/(2.00m)
= -35V/m or -35 N/C

*V/m = N/C*

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