Potential Energy of a System of Charges

\[V=\frac{1}{4\pi\varepsilon_{0}}\left[\frac{q_{1}}{r_{1}}+\frac{q_{2}}{r_{2}}+\frac{q_{2}}{r_{3}}+\frac{q_{4}}{r_{4}}+.........+\frac{q_{n}}{r_{n}}\right]\]

\[V=\frac{1}{4\pi\varepsilon_0}\sum_{i=1}^{i=n}\frac{q_i}{r_i}\]

Just as work is done to lift a ball to height h against gravity (storing gravitational PE = mgh), work must be done to bring two like charges together against their repulsive force.

• This work is not lost — it is stored as the system's electrostatic potential energy.
• If the charges are released, this stored energy converts back into kinetic energy.
• For unlike charges (attractive force), an external agent must do work to separate them. The system already has negative potential energy.

Key Physical Meaning: Electrostatic potential energy = total work done by an external agent (without acceleration) to assemble the charge configuration from infinity.

The electrostatic potential energy of a system of charges is defined as the work done by an external agent in assembling the charges at their respective positions, bringing each charge from infinity, without any kinetic energy being imparted.

• Symbol: U
• SI Unit: Joule (J)
• Nature: Scalar quantity
• Reference: U = 0 when all charges are at infinity

The electrostatic force is a conservative force — the work done in moving a charge between two points is independent of the path and depends only on the initial and final positions. This is why the potential energy is well-defined.

Setup

Consider two point charges q₁​ and q₂ separated by distance r₁₂.

Derivation (Step-by-Step)

Step 1: Bring q₁​ from infinity to Position 1. No other charge is present, so W₁ = 0.

Step 2: Bring q₂​ from infinity to Position 2, while q₁​ is already fixed at Position 1.
The electric potential at Position 2 due to q₁​ is:

Work done in bringing q₂​ to Position 2:

Step 3: Total work done = Potential Energy of the system:

Setup

Three charges q₁​, q₂​, q₃​ at positions \[\vec {r_1}\]​, \[\vec {r_2}\], \[\vec {r_3}\]:

 

Derivation

Work done at each step:

• Step 1: Bring q₁​ from ∞ → W₁ = 0
• Step 2: Bring q₂​ from ∞ in field of q₁​:
  W₂ = \[\frac{1}{4\pi\varepsilon_{0}}\cdot\frac{q_{1}q_{2}}{r_{12}}\]
• Step 3: Bring q_3​ from ∞ in field of both q₁​ and q₂​:
  W₃ = \[\frac{1}{4\pi\varepsilon_0}\left(\frac{q_1q_3}{r_{13}}+\frac{q_2q_3}{r_{23}}\right)\]

Total potential energy:

For nn charges q₁, q₂, …, q_n​ at positions r₁, r₂, …, r_n​:

Where r_ij is the distance between charge i and charge j, and the condition j > i ensures each pair is counted only once.

Given: Four charges +q, −q, +q, −q arranged alternately at corners A, B, C, D of a square of side d.

Part (a): Work to assemble the configuration

The charges are brought one by one from infinity. Work done at each step:

• Step 1: Bring +q to A: No other charge present → W₁ = 0

• Step 2: Bring −q to B (with +q at A):
  W₂ = (−q) × \[\frac {kq}{d}\] = −\[\frac {kq^2}{d}\]

• Step 3: Bring +q to C (with +q at A and −q at B). Distances: AC = d√2, BC = d:
  W₃ = (+q)\[\left(\frac{k(+q)}{d\sqrt{2}}+\frac{k(-q)}{d}\right)=\frac{kq^2}{d}\left(\frac{1}{\sqrt{2}}-1\right)\]

• Step 4: Bring −q to D (with +q at A, −q at B, +q at C). Distances: AD = d, BD = d√2, CD = d:
  W₄ = \[\left(\frac{k(+q)}{d}+\frac{k(-q)}{d\sqrt{2}}+\frac{k(+q)}{d}\right)=\frac{kq^2}{d}\left(\frac{1}{\sqrt{2}}-2\right)\]

Total work = Total potential energy U:

Part (b): Extra work to bring charge q₀ to centre E

The four charges (+q, −q, +q, −q) are placed symmetrically. The potential at the centre E due to the four charges cancels out to zero (potential from +q pairs is exactly cancelled by potential from −q pairs, since all are equidistant from E).