# Calculating Electric Potential in A Uniform Electric Field

1. Jan 21, 2004

### Moxin

Here's the problem:

Two points are in an E-field: Point 1 is at (x1,y1) = (4,4) in m, and Point 2 is at (x2,y2) = (13,13) in m. The Electric Field is constant, with a magnitude of 65 V/m, and is directed parallel to the +x-axis. The potential at point 1 is 1000 V. Calculate the potential at point 2.

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IT IS ABSOLUTELY SICKENING How Many Times I Attempted this Seemingly Easy Problem And Got it WRONG...so apparently, this isn't as easy as I thought !

I KNOW this problem Has to utilize the formula V = Ed (or perhaps V = Edcos(theta) ???)

for d i get sqrt((13-4)^2 + (13-4)^2) = 12.7279

and E is given

soooooooooo... for the change in potential i get 827. I then add that to the potential of point 1 to get the potential of point 2 and I get 1827. But apparently that's wrong. So are the answers 1000, 1585, and 1292 which I got from slightly tweaking the main formula in different ways. I have no clue what else to try...any help ?

Last edited: Jan 21, 2004
2. Jan 21, 2004

### himanshu121

First of All look at the direction of Electric field it is in +x direction So Potential is going to decrease in +x direction

Moreover Apply

$$E\cos(45)^0r=|\Delta V|$$

3. Jan 21, 2004

### HallsofIvy

Staff Emeritus
himanshu121: Do you really need to worry about the angle? Since the Field is directed in the x-direction, the potential only changes in that direction. For a constant field intensity, F, the change in potential is F times the change in x coordinate. In this problem the x coordinate changes from 4m to 13m (a change of 13-4= 9m) and F= 65 V/m so the change in potential is 65 V/m * 9 m= 585 V. As himansh121 pointed out, this is a decrease so it is -585 volts. Now, you know the potential at (4,4) so what is the potential at (13,13)?

4. Jan 21, 2004

### himanshu121

Yup, rcos45=13-4, I got ur reply Halls

the equipotential surface is a planar surface hence it would be same for x=c which is || yz

5. Jan 21, 2004

### Moxin

AHHHHHHHHHHHHHHHHH so my main problem was the sign... lol dang I gotta becareful of that, thanks.. and thanks himanshu