ABSTRACT
Consider two stationary point charges, Q1 and Q2, separated by a distance R. Let R be the distance vector from Q1 to Q2. Then Coulomb’s law states that the electrostatic force F exerted upon Q2 by Q1 is
F ¼ 1 4p«0
Q1Q2 R2
i (17:1)
where «0 is the permittivity of free space, which is a constant i is a unit vector in the direction of R
If Q1 and Q2 have the same sign, then F is positive and therefore in the direction of i (and R), meaning that Q2 is being repelled away from Q1. When Q1 and Q2 have different signs, F is negative and hence opposite to i; Q2 is now being attracted to Q1. (An analogous analysis applies to the force exerted by Q2 upon Q1)
Suppose next that Q2 is brought in infinitesimal increments dR from infinity to the separation R. The work involved is
W ¼ ðR
1 F dR ¼
ðR
1 jFj cos ujdRj (17:2)
where u is the angle between F and dR; the latter is opposite in direction to R. If F is positive (repulsive), then u¼ 1808, cos u¼1; for F negative (attractive), u¼ 08, cos u¼ 1. Substituting Equation 17.1 into Equation 17.2 and integrating yields W¼ (1=4p«0)(jQ1Q2j=R) when Q1 and Q2 have the same sign, and W¼(1=4p«0) (jQ1Q2j=R) when they have different signs. These two expressions can be combined into one by removing the absolute value signs:
W ¼ 1 4p«0
Q1Q2 R
¼ DE (17:3)
If the process of bringing Q1 and Q2 together is adiabatic (no heat transfer), then W¼DE, the change in energy. Thus energy must be provided (DE> 0) to bring Q1 and Q2 together when they repel, and it is released (DE< 0) when they attract. DE, as given by Equation 17.3, is the interaction energy of Q1 and Q2.
