The relationship between the potential and the electric field allows to find the results for the value of the electric field as a function of the distance is:
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In the attachment we see the graph of the electric field as a function of distance.
Electric potential is defined by the change in potential energy of a test charge between two points, between the value of the test charge.
dV = - E . ds
E = [tex]- \frac{dV}{ds} \ \hat s[/tex]
Where the bold letters indicate vectors, V is the potential difference, E the electric field and s the path.
Let's apply this expression for each section of the given graph:
1) section from x₀ = 0 to x_f = 2 m, the potential is V₀ = 2 V is constant.
The derivative of a constant is zero.
E = 0
2) Section between x₀ = 2 and x_f = 4 m, the potential varies linearly from V₀ = 2 v to V_f = -2 V.
We look for the equation of the line.
V-V₀ = m (x- x₀)
We carry out the derivative.
E = - m i ^
The slope (m) is:
[tex]m= \frac{V_f - V_o}{x_f- x_o}[/tex]
Let's calculate.
[tex]m= \frac{-2 -2}{4-2} = \ -2 \ V/m[/tex]
Let's substitute.
E = [tex]2 \hat i \ V/m[/tex]
3) From x₀ = 4 to x_f = 4.5 m, the potential varies from V₀ = -2 to V_f = 0.
We look for the equation of the line and we derive.
E = - m i ^
Let's substitute.
[tex]m = \frac{0-(-2)}{4.5-4} = \ 4 V/m[/tex]
E = - 4 [tex]\hat i[/tex] V / m
4) From x₀ = 4.5 m to x_f = 6m. The potential is constant and the derivative of a constant is zero.
E = 0
5) From x₀ = 6m to x_f = 8 m, the potential changes linearly from v₀ = 0 to V_f = 1 V
We look for the equation of the line and we derive.
E = - m i ^
[tex]m = \frac{1-0}{8-6} = \ 0.5 \ V/m[/tex]
E = - 0.5 [tex]\hat i[/tex] V/m
6) From x₀ = 8m to x_f = 9m, the potential changes linearly from V₀ = 1 V to V_f = -1.
We look for the equation of the line and we derive.
E = - m i ^
[tex]m = \frac{-1-1}{9-8} = \ -2 \ V/m[/tex]
Let's substitute.
E = 2 [tex]\hat i[/tex] V/m
7) From x₀ = 9m to x_f = 10 m, the potential changes linearly from V₀ = -1 V to V_f = -2 V
We look for the equation of the line and we derive.
E = - m i ^
[tex]m = \frac{-2+1}{10-9} = \ -1 \ V/m[/tex]
Let's substitute.
E = 1 [tex]\hat i[/tex] V/m
In the attachment we can see these Electric fields as a function of distance.
In conclusion, the relationship between the potential and the electric field we can find the results for the value of the electric field as a function of the distance is:
-
In the attachment we see the graph of the electric field as a function of distance.
Learn more about the electric field here: brainly.com/question/14306881
