Coulomb\'s Law yields an expression for the energy of interaction for a pair of
ID: 1430487 • Letter: C
Question
Coulomb's Law yields an expression for the energy of interaction for a pair of point charges.
V is the energy (in J) required to bring the two charges from infinite distance separation to distance r (in nm).
Q1 and Q2 are the charges in terms of electrons.
(i.e. the constant in the above expression is 2.31×10-19J 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.
d = 0.210 nm.
Explanation / Answer
You have FOUR (4) Particles to work with. You will focus on ONE CHARGE AT A TIME, and look at its interaction with EACH OF THE OTHER CHARGES. The variables Q1 and Q2 are just that: variable, meaning the number can go as high as the number of particles involved....4, so we have:
Q1 = Q3 = (-2)
Q2 = Q4 = (+2)...... I just assigned them in order from left to right in your very nice diagram.
Now, the (r) in the Coulomb equation just means "distance between the two points," and that is just you value of d, to repeat it
r = d = 5.05E-10 m
Now taking them one at a time:
V12 = energy of interaction between Q1 & Q2
= [2.31E-19(Q1)(Q2)]/r = (2.31E-19 J*m) * [(-2)(+2)/(2.10E-10 m)] = -4.4e-9 J
V13 = (2.31E-19 J*m) * [(-2)(-2)/(2 x 2.10E-10 m)] = 2.2e-9 J / 2 = 1.1e-9 J
Here (r) = 2 x (d)
V14 = (2.31E-19 J*m) * [(-2)(+2)/(3 x 2.10E-10 m)] = -1.46e-9 J / 3 = -4.88e-10 J
Here (r) = 3 x (d)
Now if you want the value for say V23, you just notice that the distance between them is the same as that for V12, the charge magnitudes are equal, and the signs are the same...so
V23 = V12
V34 = V12
Now, for V24, this will be the exact same as V13, because the distance involved in each case is 2 x d, and the magnitudes of charge are the same, and they both involve like charges.
V13 = V24
No need to do the 3 equations for Every single particle THIS TIME. But if the charges were differenct in each case, and/or the value of d was NOT constant, then we would.
So for the Total Energy of Interaction for THE SYSTEM:
V12 + V13 +V14 + V23 + V24 + V34 =
substitute according what we found above,
V12 + V13 +V14 + V12 + V13 + V12 = 3(V12) + 2(V13) + V14
3( -4.4e-9 J) + 2(1.1e-9 J) + ( -4.88e-10 J) = - 1.14e-8 J