Answer :
Answer:
400 V
Explanation:
Electric filed , E=800 V/m
[tex]V_1=2000 V[/tex]
[tex]x_1=0,x_2=2 m[/tex]
We know that
[tex]E=\frac{V_1-V_2}{x_2-x_1}[/tex]
Using the formula
[tex]800=\frac{2000-V_2}{2-0}[/tex]
[tex]800=\frac{2000-V_2}{2}[/tex]
[tex]2000-V_2=800\times 2=1600[/tex]
[tex]V_2=2000-1600[/tex]
[tex]V_2=400 V[/tex]
Hence, the potential at x=2 m=400 V
The potential difference at point x = 2m on the x-axis is 400 V.
Potential Difference
The amount of energy transferred between two points in a circuit is measured by the potential difference between two points.
Given that the electric field along the x-axis is 800 V/m. Potential difference at x= 0 is 2000 V.
The electric field at two points along the x-axis can be given as,
[tex]E = \dfrac {v_1-v_2} {x_2 -x_1}[/tex]
Where E is the electric field, v1 is the potential difference at point x1 on the x-axis, and v2 is the potential difference at point x2 on the x-axis.
The potential difference v2 at point x2 = 2 m can be calculated by substituting the values in the above equation.
[tex]800 = \dfrac {2000 -v_2}{2-0}[/tex]
[tex]v_2 = 2000 - (800\times 2)[/tex]
[tex]v_2 = 400 \;\rm V[/tex]
Hence we can conclude that the potential difference at point x = 2m on the x-axis is 400 V.
To know more about the potential difference, follow the link given below.
https://brainly.com/question/17013128.