Finding potential of two points in a constant electric field

[skibum143]

Homework Statement

The figure below shows two points in an E-field: Point 1: (3,4) and Point 2: (12,9) both in m. The electric field is constant, with a magnitude of 80 V/m, and is directed parallel to the +X axis. The potential at point 1 is 1100 V. What is the potential at point 2?

Homework Equations

E = change in V / change in X
A^2 + B^2 = C^2

The Attempt at a Solution

I got that the distance (tangent) between points 1 and 2 is 10.296 m. If I plug that into the equation to solve for V, I get 80=(V2 - 1100) / 10.296, or V2 = 823.68. Since Point 2 is further away from the origin of the E-field, the potential should be lower, so 1100 - 823.68 = 276.32. This is wrong. I tried adding them to see if that was right, (1923.68) but that was wrong too.

My main question is does the E-field value change? It says it is constant, so I don't know why it would. Thanks for the help!

The second part of the question is to calculate the work required to move a negative charge of Q = -608 microC from point 1 to point 2, which I would use the formula: the change in PE = the change in V times the charge (q). Is this right?

 

[tiny-tim]

Welcome to PF!

Hi skibum143! Welcome to PF!

(try using the X² tag just above the Reply box )

Potential = potential energy per charge …

potential energy = work done …

what is the work done if you go a distance 9 parallel to the x-axis?

what is the work done if you go a distance 5 parallel to the y-axis?

 

[skibum143]

Hi tiny-tim. Thanks so much for the response!
The work going a distance 5 parallel to the y-axis is zero, because it's on the same equipotential line.
So should I only be using 9 as my "r" instead of the tangent of 10.296?

 

[skibum143]

sorry, i meant use 9 as my change in x...

 

[skibum143]

I guess my other main question is, if they say the electric field is constant, at 80 V/m, is that the value at every point in the electric field? If not, how do you calculate the change in the electric field?

 

[skibum143]

AH! I got it! Thank you so much! :)

 

[tiny-tim]

Woohoo! ​

But, slow down in future! …

you don't usually get a quick response here, so you may as well take your time!