(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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?

2. Relevant equations

E = change in V / change in X

A^2 + B^2 = C^2

3. 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?

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# Homework Help: Finding potential of two points in a constant electric field

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