Calculating Electrostatic Repulsion in Nuclear Fission

AI Thread Summary
To calculate the electrostatic repulsion between two spheres formed from a fission event, the correct distance (d) between the centers of the spheres must be determined, which is not simply the radius. The formula F = k(q1q2/d^2) is applicable, where k is Coulomb's constant, and q1 and q2 are the charges of the protons. The user initially attempted to use a distance based on the radius but incorrectly assumed it could be doubled. A careful reevaluation of the distance between the spheres is necessary to arrive at the correct force calculation. Understanding the geometry of the situation is crucial for accurate results.
snash1057
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Homework Statement


At the point of fission, a nucleus with 98 protons is divided into two smaller spheres, each with 49 protons and a radius of 6.38x10^-15 m. What is the repulsive force pushing these two spheres apart

Hint: d ≠ the radius


Homework Equations


F= k (q1q2/d^2)


The Attempt at a Solution


if the radius isn't d then i don't know what is. i have tried multiplying the radius by 2 because there are now 2 spheres and setting it up like

F= 9E9 (49 x 1.6E-19/4.76E-29^2) but that is incorrect
 
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snash1057 said:
i have tried multiplying the radius by 2 because there are now 2 spheres
Makes sense to me.
and setting it up like

F= 9E9 (49 x 1.6E-19/4.76E-29^2) but that is incorrect
Redo this more carefully.
 
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