Question

Figure EX26.4 is a graph of $E_x$. The potential at the origin is $-50V$. What is the potential at $x=3.0m$?

Short answer

The required potential is$\mathbf{-}\mathbf{550}\mathbf{}\mathit{V}$

Step-by-step solution

Given information and formula used 

Given :

The potential at the origin : $-50V$

To find : The potential at :$x=3.0m$

Graph -

Theory used :

Potential difference is :$∆V=V_f-V_i=\int E_x·∆x$

Calculating the required the potential difference  

Using the formula to calculate potential difference, we have :

$$\begin{gathered} ∆V=V_f-V_i=\int E_x·∆x \\ \Rightarrow V_3-V_0=-{\int }_0^2{E}_x·dx-{\int }_2^{3}E·dx \\ \Rightarrow V_3+50=-200\times 2-{\int }_2^3(600-200x)dx↤\{fromgraph\} \\ \Rightarrow V_3=-400-50-{[600x-\frac{200x²}{2}]}_2^3=-450-(900-800)=-450-100 \\ =\boxed{\mathbf{-}\mathbf{550}\mathbf{}\mathit{V}} \end{gathered}$$