Energy of Interaction of Point Charges Coulomb\'s Law yields an expression for t
ID: 495243 • Letter: E
Question
Energy of Interaction of Point Charges Coulomb's Law yields an expression for the energy of interaction for a pair of point charges. V = 2.31 times 10^-19 Q_1Q_2/r V is the energy (in J) required to bring the two charges from infinite distance separation to distance r (in nm). Q_1 and Q_2 are the charges in terms of electrons. (i.e. the constant in the above expression is 2.31 times 10^-19 J nm electrons^-2) For a group of "point" charges (e.g. ions) the total energy of interaction is the sum of the interaction energies for the individual pairs. Calculate the energy of interaction for the square arrangement of ions shown in the diagram below.Explanation / Answer
Consider the number of interactions possible as shown in the rough sketch above:
1) AB = BC = CD = DA
2) AC
3) BD
Calculate the distance AC = BD (both are same since both are diagonals of a square); therefore AC = BD = (d2 + d2) = d2 = 1.414*d = 1.414*(0.545 nm) = 0.77063 nm 0.771 nm
Find out the interactions AB = BC = CD = DA (all are equivalent since all of these involve interactions between -1 and +1 charges).
AB: V = 2.31*10-19*(-1)*(+1)/(0.545) J = -4.2385*10-19 J
BC: V = 2.31*10-19*(_1)*(-1)/(0.545) J = -4.2385*10-19 J
Next consider the interactions AC and BD.
AC: V = 2.31*10-19*(-1)*(-1)/0.771 J = 2.996*10-19 J
BD: V = 2.31*10-19*(+1)*(+1)/0.771 J = 2.996*10-19 J
The total interaction energy is (4*AB + 2*AC) = [4*(-4.2385*10-19 J) + 2*(2.996*10-19 J)] = -16.954*10-19 J (ans).
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