Physics Electromagnetism - How to Solve: Electric Potential & Potential Energy (Dipole, Equipotential Surfaces) – IIT JEE Guide

How to Solve: Electric Potential & Potential Energy (Dipole, Equipotential Surfaces) – IIT JEE Guide

Introduction Mastering electric potential and dipoles unlocks 5–7 marks in IIT JEE—enough to push you from a 90th to a 99th percentile rank. These concepts also power real-world tech like MRI machines, capacitors in smartphones, and lightning rods.

WHAT YOU NEED TO KNOW FIRST

1. Electric field (E) – Vector quantity, force per unit charge.
2. Work done in electric fields – Work = Force × displacement × cosθ.
3. Potential difference (V) – Scalar, work done per unit charge to move between two points.

(If you’re shaky on these, pause and review them first.)

KEY TERMS & FORMULAS

1. Electric Potential (V)

• Definition: Work done per unit charge to bring a test charge from infinity to a point.
• Formula: [ V = \frac{kQ}{r} \quad \text{(for a point charge)} ]
• (V) = Electric potential (Volts, V)
• (k) = Coulomb’s constant ((9 \times 10^9 \, \text{Nm}^2/\text{C}^2)) MEMORISE THIS
• (Q) = Source charge (C)
• (r) = Distance from charge (m)

2. Potential Energy (U) of a System

• Definition: Work done to assemble a charge configuration.
• Formula: [ U = qV \quad \text{(for a single charge in an external potential)} ] [ U = \frac{kQq}{r} \quad \text{(for two point charges)} ]
• (U) = Potential energy (Joules, J)
• (q) = Test charge (C)
• (V) = Potential at the point (V)

3. Electric Dipole Potential

• Definition: Potential due to two equal and opposite charges separated by distance (2a).
• Formula (along axial line): [ V = \frac{kp \cos \theta}{r^2} ]
• (p = q \times 2a) = Dipole moment (C·m) MEMORISE THIS
• (\theta) = Angle between dipole axis and position vector

• (r) = Distance from dipole center (must be (r \gg a))

• Special Cases:

• Axial line ((\theta = 0°) or (180°)):
  [
  V = \frac{kp}{r^2}
  ]
• Equatorial line ((\theta = 90°)):
  [
  V = 0
  ]

4. Equipotential Surfaces

• Definition: Surfaces where potential is constant. No work is done moving a charge along them.
• Key Property: Electric field is perpendicular to equipotential surfaces.

STEP-BY-STEP METHOD

Step 1: Identify the Charge Configuration

• Is it a point charge, dipole, or multiple charges?
• For dipoles, check if the point is on the axial or equatorial line.

Step 2: Choose the Correct Formula

• Point charge? → (V = \frac{kQ}{r})
• Dipole? → (V = \frac{kp \cos \theta}{r^2})
• Potential energy? → (U = qV) or (U = \frac{kQq}{r})

Step 3: Plug in Values & Solve

• Convert all units to SI (C, m, V, J).
• For dipoles, ensure (r \gg a) (otherwise, the formula fails).

Step 4: Check for Superposition (Multiple Charges)

• If multiple charges, add potentials algebraically (scalars, not vectors!).
• Example: (V_{\text{total}} = V_1 + V_2 + V_3 + \dots)

Step 5: Relate Potential to Potential Energy

• If a charge (q) is placed in the potential (V), its potential energy is (U = qV).

Step 6: Draw Equipotential Surfaces (If Needed)

• For a point charge: Concentric spheres.
• For a dipole: Complex surfaces, but axial and equatorial lines are key.
• Field lines are always perpendicular to equipotential surfaces.

WORKED EXAMPLES

Example 1 – Basic (Point Charge Potential)

Question: Find the electric potential at a point 0.5 m from a (+3 \mu C) charge.

Solution: 1. Identify: Point charge → Use (V = \frac{kQ}{r}). 2. Convert units:
- (Q = +3 \mu C = 3 \times 10^{-6} C)
- (r = 0.5 m) 3. Plug in:
[
V = \frac{(9 \times 10^9)(3 \times 10^{-6})}{0.5} = 54 \times 10^3 \, V = 54 \, kV
] 4. Answer: (54 \, kV)

What we did and why: - Used the point charge formula because only one charge was given. - Converted (\mu C) to (C) to match SI units. - Potential is positive because the charge is positive.

Example 2 – Medium (Dipole Potential)

Question: An electric dipole has charges (\pm 2 \, nC) separated by (4 \, mm). Find the potential at a point (1 \, m) away on the axial line.

Solution: 1. Identify: Dipole → Use (V = \frac{kp}{r^2}) (axial line, (\theta = 0°)). 2. Calculate dipole moment (p):
[
p = q \times 2a = (2 \times 10^{-9}) \times (4 \times 10^{-3}) = 8 \times 10^{-12} \, C \cdot m
] 3. Plug in:
[
V = \frac{(9 \times 10^9)(8 \times 10^{-12})}{(1)^2} = 72 \times 10^{-3} \, V = 72 \, mV
] 4. Answer: (72 \, mV)

What we did and why: - Used the axial line formula because the point was along the dipole axis. - Calculated (p) first because it’s needed in the formula. - Checked (r \gg a) ((1 \, m \gg 4 \, mm)) to ensure the formula is valid.

Example 3 – Exam-Style (Superposition + Potential Energy)

Question: Two charges (+4 \, \mu C) and (-4 \, \mu C) are placed (6 \, cm) apart. A (+1 \, nC) charge is placed at the midpoint. Find: (a) The electric potential at the midpoint. (b) The potential energy of the (+1 \, nC) charge.

Solution (a): 1. Identify: Two point charges → Superposition of potentials. 2. Distance from each charge to midpoint:
[
r = \frac{6 \, cm}{2} = 3 \, cm = 0.03 \, m
] 3. Calculate (V) for each charge:
[
V_1 = \frac{kQ_1}{r} = \frac{(9 \times 10^9)(4 \times 10^{-6})}{0.03} = 1.2 \times 10^6 \, V
]
[
V_2 = \frac{kQ_2}{r} = \frac{(9 \times 10^9)(-4 \times 10^{-6})}{0.03} = -1.2 \times 10^6 \, V
] 4. Total potential:
[
V_{\text{total}} = V_1 + V_2 = 1.2 \times 10^6 + (-1.2 \times 10^6) = 0 \, V
] 5. Answer (a): (0 \, V)

Solution (b): 1. Potential energy (U = qV):
[
U = (1 \times 10^{-9})(0) = 0 \, J
] 2. Answer (b): (0 \, J)

What we did and why: - Used superposition because there were two charges. - Potential is a scalar, so we added (V_1) and (V_2) directly (no vectors!). - The net potential was zero, so the potential energy was also zero.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is your 60-second crash course for electric potential and dipoles. First, potential is a scalar, so just add it up for multiple charges. For a point charge, (V = \frac{kQ}{r})—memorise this. For a dipole, (V = \frac{kp \cos \theta}{r^2}), but only if (r) is way bigger than the dipole length. On the axial line, (\cos \theta = 1); on the equatorial line, (\cos 90° = 0), so potential is zero. Equipotential surfaces? They’re always perpendicular to field lines—no work is done moving along them. Potential energy? (U = qV). If the question gives two charges, use superposition—add potentials, not fields. Watch out for unit traps—convert everything to Coulombs and meters. And if (r) isn’t much bigger than the dipole length, don’t use the dipole formula—calculate each charge separately. You’ve got this!