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Potential Energy in an External Field - Potential Energy of a System of Two Charges in an External Field

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Estimated time: 10 minutes
CBSE: Class 12

Analogy — Gravitational Parallel

Think of two masses, m1 and m2, placed in an external gravitational field (like Earth's field). To bring them from infinity:

  • Each mass gains gravitational PE with the Earth's field → m1gh1 + m2gh2
  • They also interact with each other → mutual gravitational PE

Analogously, two charges in an external electric field gain:

  • Individual interaction PE with the external field → q1V(r1) + q2V(r2)
  • Mutual Coulombic interaction PE → \[\frac{q_1q_2}{4\pi\varepsilon_0r_{12}}\]
CBSE: Class 12

Step-by-Step Derivation

Setup

Two charges, q1 and q2, are to be placed at positions r1 and r2, respectively, in a region where an external electric field E exists, with corresponding external potential V(r).

Step 1 — Bring q1​ from infinity to r1

  • At this stage, only the external field is present. There is no other charge in the system.
  • Work done against the external field to place q1​ at r1: W1 = q1 V(r1)
  • This work is stored as potential energy of q1​ in the external field.

Step 2 — Bring q2​ from infinity to r2​

Now q1​ is already at r1. When moving q2​ to r2, work is done against two agents:

  • Agent A — External field: W2,ext = q2 V(r2)
  • Agent B — Field due to q1​:
    W2,q1 = \[\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r_{12}}\]
  • Total work done in Step 2 (by superposition of fields):
    W2 = q2 V(r2) + \[\frac{q_1q_2}{4\pi\varepsilon_0r_{12}}\]

Step 3 — Add the Work Done in Both Steps

Total potential energy = W1 + W2

U = q1V(r1) + q2V(r2) + \[\frac{q_{1}q_{2}}{4\pi\varepsilon_{0}r_{12}}\]
CBSE: Class 12

The Three Energy Components

Component Formula / Interpretation Physical Meaning
Interaction of (q1) with the external field q1 V(r1) Potential energy of charge (q1) in the external potential
Interaction of (q2) with the external field q2 V(r2) Potential energy of charge (q2) in the external potential
Mutual interaction of the two charges \[\displaystyle \frac{q_1 q_2}{4\pi \varepsilon_0 r_{12}}\] Coulomb potential energy between the two charges
CBSE: Class 12

Compare: With vs. Without External Field

Situation Potential Energy Formula
Two charges, no external field (isolated system) \[\displaystyle U=\frac{q_1q_2}{4\pi\varepsilon_0 r_{12}}\]
Single charge q in an external field U = qV(r)
Two charges q1, q2 in an external field U = \[q_1V(r_1)+q_2V(r_2)+\frac{q_1q_2}{4\pi\varepsilon_0 r_{12}}\]

The formula for two charges in an external field is simply the sum of the single-charge external-field terms, plus the mutual interaction term.

CBSE: Class 12

Sign Conventions

Condition Sign of \[\displaystyle \frac{q_1q_2}{4\pi\varepsilon_0 r_{12}}\] Physical Meaning
Like charges (q1q2 > 0) Positive Repulsive force; positive work is required to bring the charges together.
Unlike charges (q1q2 < 0) Negative Attractive force; the system releases energy as the charges come together.
CBSE: Class 12

Example

Given:

  • q1 = +7 μC = 7 × 10−6 C, located at (−9 cm, 0, 0)
  • q2 = −2 μC = −2 × 10−6 C, located at (+9 cm, 0, 0)
  • Separation: r12 = 9 + 9 = 18 cm = 0.18 m

(a) — Electrostatic PE without external field

U = \[\frac{1}{4\pi\varepsilon_0}\cdot\frac{q_1q_2}{r_{12}}=9\times10^9\times\frac{(7\times10^{-6})\times(-2\times10^{-6})}{0.18}\]
U = 9 × 109 × \[\frac{-14\times10^{-12}}{0.18}=\frac{-126\times10^{-3}}{0.18}\]
U = −0.7 J

Since q1q2 < 0, the potential energy is negative (attractive system).

(b) — Work to separate them to infinity

W = Ufinal − Uinitial = 0 − (−0.7) = +0.7 J

A positive amount of work is needed to pull apart unlike charges against the attractive Coulombic force.

(c) — PE with external field E = A/r2, A = 9 × 105 N C−1m2

Find the external potential at each charge location:

Since E = A/r2, integrating: V(r) = A/r

V(r1) = \[\frac{A}{r_1}=\frac{9\times10^5}{0.09}=10^7\mathrm{V}\]
V(r2) = \[\frac{A}{r_2}=\frac{9\times10^5}{0.09}=10^7\mathrm{V}\]

Calculate interaction energies with the external field:

q1V(r1) = (7 × 10−6) × 107 = 70 J
q2V(r2) = (−2 × 10−6) × 107 = −20 J

Apply the full formula [Eq. 2.29]:

U = q1V(r1) + q2V(r2) + \[\frac{q_1q_2}{4\pi\varepsilon_0r_{12}}\]
U = 70 + (−20) + (−0.7)
U = 49.3 J

The mutual interaction energy (−0.7 J) remains unchanged when the external field is switched on.

Test Yourself

Related QuestionsVIEW ALL [8]

For two unlike charges (q1q2 < 0), what is the physical meaning of the negative mutual Coulombic interaction energy?

Two tiny spheres carrying charges 1.5 μC and 2.5 μC are located 30 cm apart. Find the potential and electric field:

(a) at the mid-point of the line joining the two charges, and

(b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.

Consider the following statements and select the correct option.
  1. In an external electric field, the positive and negative charges of a non–polar molecule are displaced in opposite directions.
  2. In non –polar molecules displacement stops when the external force on the constituent charges of the molecule is balanced by the restoring force.
  3. The non–polar molecule develops an induced dipole moment.

Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.

Deduce the expression for the potential energy of a system of two charges q1 and q2 located `vec(r_1)` and `vec(r_2)`, respectively, in an external electric field.

Deduce the expression for the potential energy of a system of two point charges q1 and q2 brought from infinity to the points `vecr_1`and `vecr_2`respectively in the presence of external electric field `vecE`.

Two charges −q and +q are located at points (0, 0, −a) and (0, 0, a), respectively.

(a) What is the electrostatic potential at the points?

(b) Obtain the dependence of potential on the distance r of a point from the origin when r/a >> 1.

(c) How much work is done in moving a small test charge from the point (5, 0, 0) to (−7, 0, 0) along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?

What is the primary analogy between gravitational potential energy and electric potential energy for a system of two masses/charges in an external field?

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