Calculating Electric Potential in A Uniform Electric Field

In summary, the problem involves two points in an E-field with a constant magnitude of 65 V/m and directed parallel to the +x-axis. Point 1 is at coordinates (4,4) in m with a potential of 1000 V, and Point 2 is at coordinates (13,13) in m. The task is to calculate the potential at Point 2. Through utilizing the formula V = Ed, the potential at Point 2 is found to be -585 V, as the direction of the electric field results in a decrease in potential in the +x direction. The potential at Point 1 is used as a reference for the potential at Point 2.
  • #1
Moxin
24
0
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 ?
 
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  • #2
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

[tex]E\cos(45)^0r=|\Delta V|[/tex]
 
  • #3
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
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
Originally posted by HallsofIvy
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)?

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

1. What is an electric potential in a uniform electric field?

Electric potential is a measure of the amount of work needed to move a unit of positive charge from one point to another in an electric field. In a uniform electric field, the electric potential is constant at all points.

2. How do you calculate the electric potential in a uniform electric field?

The electric potential in a uniform electric field can be calculated using the formula V = Ed, where V is the electric potential, E is the electric field strength, and d is the distance between the two points.

3. Can the electric potential in a uniform electric field be negative?

Yes, the electric potential in a uniform electric field can be negative. This indicates that the potential energy of a positive charge is decreasing as it moves in the direction of the electric field.

4. What is the unit of electric potential in a uniform electric field?

The unit of electric potential in a uniform electric field is volts (V). It is defined as 1 joule of work per 1 coulomb of charge moved.

5. How does the distance between two points affect the electric potential in a uniform electric field?

The electric potential in a uniform electric field is directly proportional to the distance between two points. This means that as the distance increases, the electric potential also increases. In other words, the electric potential decreases as you move away from the source of the electric field.

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