Question

25.Figure 22.22 shows a plot

Answer 1

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:

• 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 = $- \frac{dV}{ds} \ \hat s$  

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:

       $m= \frac{V_f - V_o}{x_f- x_o}$  

Let's calculate.

       $m= \frac{-2 -2}{4-2} = \ -2 \ V/m$  

Let's substitute.

       E =  $2 \hat i \ V/m$  

         

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.

      $m = \frac{0-(-2)}{4.5-4} = \ 4 V/m$  

    E = - 4 $\hat i$ 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 ^

       $m = \frac{1-0}{8-6} = \ 0.5 \ V/m$  

      E = - 0.5 $\hat i$ 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 ^

       $m = \frac{-1-1}{9-8} = \ -2 \ V/m$

Let's substitute.

       E = 2 $\hat i$ 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 ^

       $m = \frac{-2+1}{10-9} = \ -1 \ V/m$

Let's substitute.

       E = 1 $\hat i$  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