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How large in N is the electrostatic repulsion between two protons a distance d = 6.7 fm apart in a nucleus?
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whats N? newton?
F=kQQ'/r^2..Q and Q' are the two charges charge of a proton here... r is the distance between them( strictly speaking between their centres) k is a constant: 8.98*10^9
btw Q=Q' here..so u can write Q^2....Q being the charge on a proton
do you know what the charge would be?
ohhh nvm!
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if i recall correctly..its 1.62*10^-19C
So I did F=8.98E9*(1.6E-19^2)/(6.7^2)=5.12E-30 but it says its wrong
6.7 fm sweetheart...its 6.7*10^-15 i guess
wats the answer?
@witthan1
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One sec sorry
5.12 thank you!
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