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Physics · Ch 2 — Electrostatic Potential and Capacitance

Potential Energy in an External Field

2.8

Potential Energy in an External Field

Potential Energy in an External Field

When a charge is placed in an external electric field, it experiences a force. To bring the charge from infinity (where the field is zero) to a specific point in the field, work must be done against this force. This work gets stored as potential energy of the charge in the external field.


1. Potential Energy of a Single Charge qq in an External Field

Consider a point charge qq placed in an external electric field E\mathbf{E} (produced by other charges not included in the system). The electrostatic potential at a point in this field is defined as the work done in bringing a unit positive charge from infinity to that point.

  • If the potential at a point is VV, then the potential energy UU of a charge qq placed there is:

U=qVU = q V

  • Meaning: UU is the work done by an external agent in bringing the charge qq from infinity to that point, without accelerating it.
  • Sign convention: If qq and VV have the same sign, UU is positive (work is done against the field). If opposite signs, UU is negative (work is done by the field).

2. Potential Energy of a System of Two Charges in an External Field

Now consider two point charges q1q_1 and q2q_2 placed at positions r1\mathbf{r}_1 and r2\mathbf{r}_2 in an external field E\mathbf{E}. The total potential energy of the system is the sum of:

  1. The work done to bring q1q_1 from infinity to r1\mathbf{r}_1 in the external field.
  2. The work done to bring q2q_2 from infinity to r2\mathbf{r}_2 in the combined field of the external source and q1q_1.

Step-by-step derivation:

  • Step 1: Bring q1q_1 from infinity to r1\mathbf{r}_1. The work done is:

W1=q1V(r1)W_1 = q_1 V(\mathbf{r}_1)

where V(r1)V(\mathbf{r}_1) is the external potential at r1\mathbf{r}_1.

  • Step 2: Now bring q2q_2 from infinity to r2\mathbf{r}_2. The external agent must work against:

    • The external field (potential V(r2)V(\mathbf{r}_2)).
    • The field of q1q_1 (potential at r2\mathbf{r}_2 due to q1q_1 is 14πϵ0q1r12\frac{1}{4\pi\epsilon_0} \frac{q_1}{r_{12}}, where r12=∣r2−r1∣r_{12} = |\mathbf{r}_2 - \mathbf{r}_1|).

    The work done in this step is:

W2=q2V(r2)+14πϵ0q1q2r12W_2 = q_2 V(\mathbf{r}_2) + \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}}

  • Total potential energy UU of the two-charge system in the external field is:

U=W1+W2=q1V(r1)+q2V(r2)+14πϵ0q1q2r12U = W_1 + W_2 = q_1 V(\mathbf{r}_1) + q_2 V(\mathbf{r}_2) + \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}}

  • Key insight: The term 14πϵ0q1q2r12\frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}} is the mutual potential energy of the two charges — it does not depend on the external field. It is the work done to assemble the two charges in the absence of any external field.

3. Generalization to nn Charges …