Question

Energy of Interaction of Point Charges Coulomb's Law yields an expression for the energy of interaction for a pair of point charges. 2.31 x 10-19Q1Q2 V is the energy (in )) required to bring the two charges from infinite distance separation to distance r (in nm). Qi and Q2 are the charges in terms of electrons. (i.e. the constant in the above expression is 2.31x10-19 nm electrons") 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 linear arrangement of ions shown in the diagram below. d = 0.640 nm. -2.89*10^-18J Make sure to include ALL pairs and get the signs correct. The energy is positive for repulsive interactions, whereas for attractive interactions the energy is negative.

Answer 1

bio V = 2.31 x 10-19 x Qi Q2 42 d- kd med id=0.640 CM par fisust þair AB] = 0,7-2 V = 3:31X1019 xfQx(0) VP-1644X10-13 J = +2 0.640 I similarly for [B-G) and (c-D Pair because the charges tuumain sam so total eningy Ve 3.X-1144 X10-18 ] V=-4.32 X10-18 # for pain A-cl_and | B-D) va dr31x10 " x tabylta) Va:Slx 19 x (+a)= (42 2x0,67 B-D Luas +1.44 x 10+193) + Voi Vare + Voe mes so total Energy - V, + V2 = 4.32x10 18 J+1997]ölby V = 2.89 x10-18 2x0,67