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:
- (190V-120V)/(5.00m-3.00m)
= - (70V)/(2.00m)
= -35V/m or -35 N/C

*V/m = N/C*

To find the electric field in this region, we can use the formula:

Electric field (E) = Change in electric potential (ΔV) / Change in position (Δx)

Given the electric potentials at x = 3.0 m and x = 5.0 m, we can calculate the change in electric potential as:

ΔV = V2 - V1
ΔV = 190 V - 120 V
ΔV = 70 V

Likewise, we can calculate the change in position as:

Δx = x2 - x1
Δx = 5.0 m - 3.0 m
Δx = 2.0 m

Now, we can substitute the values into the formula to find the electric field:

E = ΔV / Δx
E = 70 V / 2.0 m
E = 35 V/m

Therefore, the electric field in this region, assuming it's constant, is 35 V/m.

To find the electric field in the given region, we can use the formula:

E = ΔV / Δx

where E is the electric field, ΔV is the change in electric potential, and Δx is the change in position.

In this case, ΔV = (190 V - 120 V) = 70 V (the change in electric potential)
and Δx = (5.0 m - 3.0 m) = 2.0 m (the change in position)

Now, we can substitute the values into the formula:

E = (70 V) / (2.0 m) = 35 V/m

Therefore, the electric field in this region, assuming it's constant, is 35 V/m.