Why is the percentage uncertainty in the electric field larger than the percentage uncertainty in the first row of voltages? Hint: Is the difference in voltage smaller than the voltages in the first row?

ΔE / E = ΔV / V + Δd / d

Explanation:

The electric field and voltage are related by the expression

      E = V / d

The uncertainty of the field is

       ΔE / E = ΔV / V + Δd / d

We see that the uncertainty of the field must be added the uncertainty of the distance that in the measure of the different potential has no effect.

The only case in which the uncertainties are equal is if the distance measurement has no uncertainty, or the separation of very high

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