Answer:
24.32 V/m
Explanation:
We are given that
[tex]V_1=4 V[/tex]
[tex]V_2=3.1 V[/tex]
The distance between the points A and B on two equipotential lines=d=[tex]x_2-x_1=3.7 cm=3.7\times 10^{-2} m[/tex]
By using [tex]1 cm=10^{-2}m[/tex]
We have to find the average electric field in (volt/m) at the mid-point C expressed to one decimal places.
We know that the electric field=[tex]E=-\frac{dV}{dx}=-\frac{V_2-V_1}{x_2-x_1}[/tex]
Using the formula
Electric field=E=[tex]-\frac{(3.1-4)}{3.7\times 10^{-2}}[/tex]
[tex]E=24.32 V/m[/tex]
