Asked by Cal
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?
Answers
Answered by
drwls
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
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
Answered by
ant
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*
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*
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.