Answer:
The electric field at that point is [tex]E = 66.66 \Delta V[/tex]
Explanation:
From the question we are told that
The distance is [tex]d = 1.5 cm = \frac{1.5}{100} = 0.015 \ m[/tex]
Generally the electric field at this region is mathematically represented as
[tex]E = \frac{\Delta V }{d}[/tex]
So [tex]\Delta V[/tex] is the electric potential difference between the electrods
So the equation becomes
[tex]E = \frac{\Delta V }{0.015}[/tex]
[tex]E = 66.66 \Delta V[/tex]
When the value of potential difference from the manual is substituted for
[tex]\Delta V[/tex] is substituted the we would arrive at the value of E but these is not available([tex]\Delta V[/tex]) i will leave the solution in terms of [tex]\Delta V[/tex]
