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Q.An electric dipole consisting of charges +q+q and −q-q separated by a distance LL is in stable equilibrium in a uniform electric field E⃗\vec{E}. The electrostatic potential energy of the dipole is (A) qLEqLE (B) zero (C) −qLE-qLE (D) −2qEL-2qEL

CBSECBSE Class XII Board 2020MCQ· 1mImportance★★★★★
✓ Free question

In stable equilibrium, the dipole moment p⃗\vec{p} aligns with the electric field E⃗\vec{E}, giving the minimum potential energy U=−p⃗⋅E⃗=−qLEU = -\vec{p} \cdot \vec{E} = -qLE. The correct option is (C).

Potential Energy of a Dipole — The Concept First

An electric dipole placed in a uniform field experiences a torque that tries to align its dipole moment p⃗\vec{p} with the field E⃗\vec{E} — much like a compass needle turning to align with a magnetic field. The stable equilibrium position is when the dipole points along the field direction.

The electrostatic potential energy of a dipole in a uniform field is not just any number — it depends on the angle between the dipole moment and the field. The formula is:

U=−p⃗⋅E⃗=−pEcos⁡θU = -\vec{p} \cdot \vec{E} = -pE\cos\theta

where p⃗=qL⃗\vec{p} = q\vec{L} (from −q-q to +q+q), and θ\theta is the angle between p⃗\vec{p} and E⃗\vec{E}.

Step-by-Step Reasoning

  1. Identify the dipole moment.

    The dipole moment vector p⃗\vec{p} points from the negative charge to the positive charge, with magnitude p=qLp = qL. So p⃗=qL⃗\vec{p} = q\vec{L}.

  2. Recall the potential energy formula.

    For any dipole in a uniform electric field, the potential energy is U=−p⃗⋅E⃗U = -\vec{p} \cdot \vec{E}. This is derived from the work done to rotate the dipole from a reference angle (usually θ=90∘\theta = 90^\circ, where U=0U=0) to its current orientation.

  3. Understand "stable equilibrium".

    Stable equilibrium means the system returns to that position if slightly disturbed. For a dipole, this happens when the torque τ⃗=p⃗×E⃗\vec{\tau} = \vec{p} \times \vec{E} is zero and the energy is at a minimum.

    • Torque is zero when p⃗\vec{p} is parallel or antiparallel to E⃗\vec{E}.
    • The energy U=−pEcos⁡θU = -pE\cos\theta is minimum when cos⁡θ\cos\theta is maximum, i.e., cos⁡θ=+1\cos\theta = +1, which means θ=0∘\theta = 0^\circ — the dipole points along the field.
  4. Plug in the numbers.

    At θ=0∘\theta = 0^\circ, cos⁡0=1\cos 0 = 1, so:

U=−pE⋅1=−qLEU = -pE \cdot 1 = -qLE

Watch out

A common mistake is to think stable equilibrium means the dipole is antiparallel (θ=180∘\theta = 180^\circ). That gives U=+qLEU = +qLE, which is a maximum — that's unstable equilibrium, like a pencil balanced on its tip. Stable means minimum energy, so the dipole must align with the field.

  1. Check the options.
    • (A) qLEqLE — that's the unstable case.
    • (B) zero — that's when θ=90∘\theta = 90^\circ, not equilibrium.
    • (C) −qLE-qLE — matches our result.
    • (D) −2qEL-2qEL — would require p=2qLp = 2qL, which isn't the case here.
Tip

You can also think of it this way: In stable equilibrium, the system has done work to align itself, so it has given up energy — hence the negative sign. The magnitude qLEqLE is the work needed to rotate it to the perpendicular position.

✓Final answer

The correct option is (C) −qLE-qLE.

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