Coulomb\'s Law yields an expression for the energy of interaction for a pair of

Question

Coulomb's Law yields an expression for the energy of interaction for a pair of point charges. V = 2.31 times 10^-19 Q_1 Q_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

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.

Coulomb's Law : V (in J) = [ 2.3110^-19 (Q1 Q2) ] / r

Given : r = 0.675 nm

the distance (r) between +1and -1 is 0.675 nm

Q1 = Q3 = (-1)

Q2 = Q4 = (+1)...... I just assigned them in order from left to right in your very nice diagram. Which is an excellent way to do this problem

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 = 6.75E-10 m .....im expressing the distance in meters.

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) * [(1)(-1)/(6.75E-10 m)] -9 = -1.48e-9 J

V13 = (2.31E-19 J*m) * [(-1)(-1)/(2 x 5.05E-10 m)] = +9.95e-10 J

Here (r) = 2 x (d)

V14 = (2.31E-19 J*m) * [(-1)(+1)/(3 x 5.05E-10 m)] = -6.6e-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( -1.48e-9 J ) + 2(+9.95e-10 J ) + ( -6.6e-10 J) = - 8.86e-10 J

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