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Hello.
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So we need to find the magnitude of the electrostatic force between two protons located on the opposite sides of the nucleus diameter.
00:09
Now, the electrostatic force between two charges is given by kulam's law, where we know that f is equal to k of e, q1, q2, over r squared, where our k of e is kulam's constant.
00:24
This is 8 .98 times 10 to the 9th newton's times meter square over coulum square.
00:34
We also have q1, which is equal to q2, which is 1 .602 times 10th to negative 19th.
00:43
This is the charge of a proton, and r is the distance between the protons.
00:47
Now, in this case, the distance between the two protons is twice the radius, since they are on opposite sides of the nucleus...