posted by Cal on .
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?
Please show some work of your own. E-Field = -(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
exactly what drwls said.
Electric field =
-(change in charge)/(change in position)
So in this case in particular, you have:
= - (70V)/(2.00m)
= -35V/m or -35 N/C
*V/m = N/C*