Question

A graph of the x component of the electric field as a function of x in a region of space is shown in Fig. 24-35. The scale of the vertical axis is set by E_xs = 20.0 N/C. The y and z components of the electric field are zero in this region. If the electric potential at the origin is 10 V, (a) what is the electric potential at x = 2.0 m, (b) what is the greatest positive value of the electric potential for points on the x axis for which $\mathbf{0}\mathbf{\le }\mathit{x}\mathbf{\le }\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}$, and (c) for what value of x is the electric potential zero?

Short answer

1. The electric potential is V = 30 V.
2. The greatest positive value of the electric potential is V_max = 40 V .
3. The value of x where the electric potential is zero is 5.5 m.

Step-by-step solution

Given data:

The x-component of the electric field in a region of space is E_xs = 20.0 N/C.

The electric potential at the origin is V_i = 10 V.

Understanding the concept:

Using the equation of potential difference

V = Ed

Here, V is the potential difference, E is the electric field, and d is the distance between two points.

You can find the potential difference between two points by using the above equation.

(a) Calculate the electric potential at x = 2.0 m :

By using the equation of the electric potential difference between two points i and f is

$\begin{array}{rcl}{V}_f-V_i& =& -\underset{i}{\overset{f}{\int }}E·ds\\ & =& A\end{array}$

The change in potential is the negative of the “area” under the curve. Thus, calculate the area of the curve.

$\begin{array}{rcl}A& =& \frac{1}{2}\left(2.0m\right)\left(-20N/C\right)\\ & =& -20N·m/C\end{array}$

As the area is equal to the change in the potential. Thus, using the area-of-a-triangle formula, you have

$\begin{array}{rcl}V-10V& =& -\left(-20N·m/C\right)\\ V& =& 20V+10V\\ V& =& 30V\end{array}$

Hence, the electric potential is 30 V.

(b) Calculate the greatest positive value of the electric potential for points on the x axis for which 0≤x≤6.0 m :

For any region within $0<x<3m,-\int E·ds$is positive, but for any region for which x > 3m , it is negative.

Therefore, V = V_max occurs at x = 3 m .

$V_{max}-10=-\underset{0}{\overset{x=3}{\int }}E.ds$

Thus, calculate the area of the curve.

$\begin{array}{rcl}A& =& \frac{1}{2}\left(3.0m\right)\left(-20N/C\right)\\ & =& -30N·m/C\end{array}$

As the area is equal to the change in the potential. Thus, using the area-of-a-triangle formula, you have

$\begin{array}{rcl}{V}_{max}-10V& =& -\left(-30N·m/C\right)\\ V_{max}& =& 30V+10V\\ V_{max}& =& 40V\end{array}$

Hence, the maximum electric potential is 40 V.

(c) Calculate for what value of x is the electric potential zero

In view of our result in part (a), you see that now (to find V = 0 ) you are looking for some

x > 3 such that the ‘’area’’ from x = 30 to x = X is 40 V.

Using the formula for a triangle (3 < x < 4) and a rectangle (4 < x < X) , you require the total area as below.

$\begin{array}{rcl}A& =& \frac{1}{2}\left(1.0m\right)\left(20.0N/C\right)+\left(X-4.0m\right)\left(20.0N/C\right)\\ & =& 10.0N·m/C+\left(X-4.0m\right)\left(20.0N/C\right)\end{array}$

Now from part (b) you can say that the electric potential in the region 3 m < x < X is 40.0 V . So, you can write the above equation as,

$\begin{array}{rcl}10.0N·m/C+\left(X-4.0m\right)\left(20.0N/C\right)& =& 40.0V\\ \left(X-4.0m\right)\left(20.0N/C\right)& =& 40.0V-10.0V\\ \left(X-4.0m\right)& =& \frac{30.0V}{\left(20.0N/C\right)}\\ X& =& 1.5m+4.00m\\ & =& 5.5m\end{array}$

Hence, the value of x where the electric potential is zero is 5.5 m.