When mapping the equipotentials on the plates with different electrode configurations you may find that some have significant areas with uniform distribution of the equipotential lines. if the distance between such lines is 0.5 cm, what is the electric field there (in units si)? assume the applied potential difference between the electrodes is as recommended in the manual.

v = 10 V and E = 2 10³ N/C

Explanation:

The electrical potentials and the electric field at one point are related by the expression

            ΔV = - ∫ E. dS

Where the bold indicates vector quantities, E is the electric field and S is the line of displacement of the load, in general displacement is perpendicular to the equipotential lines, which reduces the product scales to the ordinary product.

 If the potential difference is the most usual that is V = 10 V, the electric field is

   s = 0.5 cm = 0.5 10⁻² m

                E = ΔV / S

                E = 10/0.5 10⁻²

                 E = 2 10³ N / C

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2