Which contribution added to Rutherford's discovery of a positive nucleus?
2 answers:
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Answer:
E = 8.5 * 10^6 V/m
Explanation:
In general we have the following relation between the Electric Field and the Elecric Potential:
Due to the vector nature of the electric filed, we can only know the mean Electric field E across the membrane, and take it out from the integral, that is:
E = (ΔV)/L
Where L is the thickness of the membrane and ΔV is the potential difference.
Therefore:
E = 8.53933*10^6 V/m
rounding to the first tenth:
E = 8.5 * 10^6 V/m