When you calculate the electrostatic potential energy of a pair of charged particles, it is enough to consider the energy of one particle in the electric field of the other. If you have three charged particles (as for example A, B & C), you will have three pairs (AB, AC & BC) and the total energy will be the algebraic sum of the energies due to these three pairs. But, if you have four charged particles (A, B, C & D), you will have six pairs (AB, AC, AD, BC, CD & BD).
Consider the following MCQ which appeared in KEAM (Engineering) 2007 question paper:
The electrostatic potential energy between proton and electron separated by a distance 1 Ǻ is
(a) 13.6 eV (b) 27.2 eV (c) 14.4 eV (d) 1.44 eV (e) 28.8 eV
This is a simple question since we have just one pair of charges. The electrostatic potential energy (EPE) of the system is (1/4πε0)(q1q2/r) where q1 and q2 are the charges of proton and electron.
Therefore, EPE = (1/4πε0)(1.6×10–19)2/10–10 joule
= 9×109×1.6×10–19/10–10 electron volt
= 14.4 eV.
[When you find the EPE of more than two charged particles, you will have to be careful to consider the sign of the charges since the total EPE is the algebraic sum of the EPE’s of all the pairs of charges].
The following simple MCQ also is from the KEAM (Engineering) 2007 question paper:
The work done in moving an alpha particle between two points having potential difference 25 volt is
(a) 8×10–18 J (b) 8×10–19 J (c) 8×10–20 J (d) 8×10–10 J (e) 4×10–18 J
This question is intended to test your basic understanding of the work done in moving a charge in an electric field. An electric field will exist only if there is a potential gradient and the wok done in moving a charge is the product of the charge and the potential difference. The alpha particle has two protons in it, each carrying positive charge of 1.6×10–19 coulomb.
Therefore work done = (2×1.6×10–19)×25 joule = 8×10–18 J
Imagination is more important than knowledge.– Albert Einstein
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