29: Problem 29.4

INTRO: The figure is a graph of E_x. The potential at the origin is -250 V

PART A: 
What is the potential at x = 3.0 m?
Express your answer to two significant figures and include the appropriate units.

SOLUTION: 
To determine the potential at the given point, we need to determine the potential difference and add the initial potential. V_f=ΔV + V_i
To get the potential difference we need to integrate the function in the graph 
ΔV =- ∫_xi^x_f E(x)dx 
since the function E(x) changes at 2, we need to integrate two different functions and put them together
The function when 0≤x≤2 is simply y=200
the function when 2≤x≤3 is y=-200x+600
the integral for 0≤x≤2 is - ∫₀² 200 dx
the integral for 2≤x≤3 is - ∫₂³ -200x+600 dx
Now to integrate the two of those,
1) - ∫₀² 200 dx = -200x |₀² = -400-0 = -400 V
2) - ∫₂³ -200x+600 dx = 100x²-600x |₂³
= 900-1800-(400-1200) = -900+800 = -100 V
ΔV = -400 V + -100 V = -500 V
V_f=ΔV + V_i = -500 V + -250 V = -750 V