... An electron is placed at each corner of a square of side L= 2.50 nm. nm= 10^
ID: 1281421 • Letter: #
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An electron is placed at each corner of a square of side L= 2.50 nm. nm= 10^-9 m (a) Find the total electric potential energy Ui of the four electrons. Note that there are six pairs of electrons to be added. Assume the electrons are released from rest simultaneously. Because of the symmetry of the configuration, they would maintain a square geometry as they repel each other. (b) Find their electric PE Uf at the instant the square has side L= 10.0 nm. (c) Find the speed of each electron in (b). They would have the same speed (d) Find the speed of each electron as L approaches infinity.Explanation / Answer
Let a be the side of the square.
Potential energy=4ke^2/a+2ke^2/sqrt(2)a=(4+sqrt(2))ke^2/a
a)U=(4+sqrt(2))*9*10^9*(1.6*10^-19)^2/(2.5*10^-9)=4.99*10^-19J
b)U=(4+sqrt(2))*9*10^9*(1.6*10^-19)^2/(10*10^-9)=1.25*10^-19J
c)Change in potential energy=gain in kinetic energy=4*0.5mv^2=3.76*10^-19J
2*9.1*10^-31*v^2=3.76*10^-19J
Solving for v,
v=4.54*10^5m/s
d)Here, all potential energy of the system converts to potential energy.
2*9.1*10^-31*v^2=4.99*10^-19J
v=5.24*10^5m/s
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