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Vikki [24]
3 years ago
6

What is the repulsive electrical force between two protons 4.0×10−15m apart from each other in an atomic nucleus?

Physics
2 answers:
Annette [7]3 years ago
5 0
F = K\frac{q1 q2}{r^2}

F =  1/(4 * 3.14 * 8.85 * 10^-12)   [ (1.6x 10^-19)^2/(4 x 10 ^-15)^2 ]  = 14.394 N
otez555 [7]3 years ago
5 0

Answer:

F = 14.4 N

Explanation:

As we know that the force between two charge particles is given by coulombs law as

F_e = \frac{kq_1q_2}{r^2}

here we have

q_1 = q_2 = e

also we have

r = 4 \times 10^{-15} m

now we will have

F = \frac{(9\times 10^9)(1.6 \times 10^{-19})(1.6 \times 10^{-19})}{(4 \times 10^{-15})^2}

F = 14.4 N

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