Questions: Electric Potential Energy in Charge Systems

Question 1

For a system of three charges, you compute q₁V₁ + q₂V₂ + q₃V₃, where Vᵢ is the total potential at charge i due to the other two charges. Why must you multiply this sum by ½ to get the correct potential energy?

• A. Because Vᵢ includes the charge's own self-energy, which must be halved to remove it
• B. Because each pairwise interaction U₁₂ appears once in q₁V₁ and once in q₂V₂ — the raw sum counts every pair twice
• C. Because the ½ reflects that only half the energy is stored in the electric field; the rest is kinetic
• D. Because the formula applies only to point charges, and ½ corrects for the extended-charge approximation

Answer: (B)

When you compute q₁V₁, the potential V₁ at charge 1's location includes contributions from charges 2 and 3 — so q₁V₁ contains the interaction energy U₁₂ and U₁₃. Similarly, q₂V₂ contains U₁₂ and U₂₃ again. Every pairwise interaction is counted twice in the raw sum. The factor of ½ corrects for this double-counting. The same reason underlies the ½ in the capacitor energy formula U = ½CV².

Question 2

A capacitor is charged to voltage V with charge Q on its plates. A student argues that the stored energy should be QV (charge times voltage), not ½QV. What physical reasoning explains the factor of ½?

• A. Half the energy is stored in the electric field and half remains on the plates as surface energy
• B. The voltage across the capacitor builds from 0 to V during charging; the average voltage during charge transfer is V/2, so total work done is Q × (V/2) = ½QV
• C. The factor of ½ comes from the quantum mechanical zero-point energy of the capacitor
• D. Energy is lost to resistance during charging, so only half of QV is actually stored

Answer: (B)

As charge accumulates on the capacitor, the voltage increases proportionally from 0 to the final value V. Each infinitesimal charge dq transferred must be moved against the current voltage q/C, so dW = (q/C)dq. Integrating from 0 to Q gives W = Q²/(2C) = ½QV = ½CV². The factor ½ is not a correction or loss — it reflects that the potential is not constant during charging, so you cannot simply multiply the final Q by the final V. The same ½ arises from the double-counting argument in the assembly formula.

Question 3

Assembling two positive charges from infinite separation to a finite distance releases energy, because like charges naturally attract.

• T. True
• F. False

Answer: False

Like charges repel. Work must be done against the repulsive force to bring two positive charges closer together — this requires energy input, not energy release. The potential energy U = kq₁q₂/r is positive for like charges, reflecting stored energy that would be released if the charges flew apart to infinity. Opposite charges attract: assembling them releases energy (U < 0), and energy must be supplied to separate them. The sign of U directly encodes this: positive U = energy was stored (forced together against repulsion), negative U = bound configuration (held together by attraction).

Question 4

For a single charge q placed at a location where the electric potential is V, the potential energy is U = qV with no factor of ½.

• T. True
• F. False

Answer: True

The factor of ½ only appears when summing qᵢVᵢ over a system of multiple interacting charges, to correct for double-counting each pair. For a single test charge q at a location where an external potential V has already been established, there is no double-counting: the charge interacts with the external field, and the energy is simply U = qV. The ½ is not a universal factor — it is a correction for a specific counting issue in the multi-charge assembly formula.

Question 5

Explain why the formula for total potential energy of a system of charges is U = ½ΣᵢqᵢVᵢ rather than ΣᵢqᵢVᵢ, and identify at least one other formula in electrostatics that contains a ½ for the same reason.

Model answer: The raw sum ΣᵢqᵢVᵢ counts each pairwise interaction twice: the interaction between charges i and j appears in both qᵢVⱼ and qⱼVᵢ. Dividing by 2 corrects for this. The same factor appears in the capacitor energy U = ½CV² = ½QV: as charge builds up on a capacitor, each increment dq is transferred at voltage q/C, and integrating gives the ½. The factor also appears in the energy density of the electric field, u = ½ε₀E², for the same underlying reason — all three formulas count the self-consistent assembly of a charge distribution.

A useful mnemonic: whenever you sum over all charges or fields in a self-consistent system (where each element contributes to the field felt by every other), a factor of ½ appears to avoid double-counting. It shows up throughout both electrostatics and magnetostatics (magnetic energy density u = B²/(2μ₀), inductor energy U = ½LI²).