Question

When mapping the equipotentials on the plates with different electrode configurations you may find that there are significant regions with uniform distribution of the equipotential lines. If the distance between such lines is 1.5 cm, what is the electric field there (in units SI)? Assume the applied potential difference between the electrodes is as recommended in the manual.

Answer 1

Answer:

The electric field at that point is $E = 66.66 \Delta V$

Explanation:

From the question we are told that

   The distance is  $d = 1.5 cm = \frac{1.5}{100} = 0.015 \ m$

Generally the electric field at this region is mathematically represented as

                 $E = \frac{\Delta V }{d}$

So   $\Delta V$  is the electric potential difference between the electrods  

   So the equation becomes

                 $E = \frac{\Delta V }{0.015}$

                  $E = 66.66 \Delta V$

When the value of  potential difference from the manual is substituted for

     $\Delta V$ is substituted the we would arrive at the value of E but these is not available($\Delta V$)  i will leave the solution in terms of  $\Delta V$