Physics, 21.02.2020 23:50, Svetakotok
Two identical conducting spheres, A and B, carry equal charge. They are stationary and are separated by a distance much larger than their diameters. A third identical conducting sphere, C, is uncharged. Sphere C is first touched to A, then to B, and finally removed (to a far away distance). If before the sphere C is introduced the electrostatic force between A and B is Fi , how does Ff , the electrostatic force between A and B after C is removed, compare to Fi?
Answers: 3
![F_i = \frac{1}{4\pi\epsilon_0}\frac{q_Aq_B}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{Q^2}{r^2}[/tex}The electrostatic force, Ff, in the final configuration is [tex]F_f = \frac{1}{4\pi\epsilon_0}\frac{q_Aq_B}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{3Q^2/8}{r^2}[/tex}Therefore, the relation between Fi and Ff is as follows[tex]F_i = F_f\frac{3}{8}\\8F_i = 3F_f](/tpl/images/0519/8332/266cb.png)
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Two identical conducting spheres, A and B, carry equal charge. They are stationary and are separated...
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