The lead nucleus has a diameter of 14.2 fm. What is the electric field strength at the surface of a lead nucleus

E = kQ / r^2 = 2.34*10^21 V/m

where k = 8.99*10^9 m/F

Q = 82 * 1.6*10^-19 C

r = 7.1*10^-15 m

Explanation:

14.2 FM = 2.34×10^21 V/m

5 w for 1hr = 5x60x60 joules=300x60=3000x6=18,000j as total energy conversion. efficiency of bulb needed for radiant to thermal within 18000

i just did an experiment like this in my science class. also the answer is yes, it will get separated because they are two opposites meaning if they are opposite they can not attract.

hope this .!