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?

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.

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.

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.

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.

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.