Electric potential: a point charge and a dipole

Potential and potential difference

Water flows from a higher level to a lower one, and heat flows from a hotter body to a colder one. Electric potential plays the same part for charge: a positive charge moves from higher to lower potential. To move a charge against the field, an outside agent must do work. This work is stored as potential energy. The potential difference between A and B is the work done per unit positive charge to move a charge from A to B. So V(B) − V(A) = W/q₀. The electrostatic field is conservative, so this work does not depend on the path. The potential at a point is the work done to bring a unit positive charge there from infinity, where V = 0. Potential is a scalar, measured in volts: 1 V = 1 J C⁻¹. For several charges, simply add their potentials: V = V₁ + V₂ + V₃ + ....

Derivation: potential due to a point charge

A point charge +q is at O. Find the potential at P, where OP = r. Bring a test charge q₀ from infinity to P in many tiny steps, through points C, B and A.

1. 1
   Take one tiny step from A to P, where OA = r_A is only a little more than r.
2. 2
   The field changes slightly over this step. Take its average value as the geometric mean: E = kq/(r_A r).
3. 3
   Work done against the field in this step: W = q₀E(r_A − r) = kqq₀(r_A − r)/(r_A r) = kqq₀(1/r − 1/r_A).
4. 4
   Do the same for every earlier step: from B to A, from C to B, and so on, all the way out to infinity.
5. 5
   Add them all. Each middle term cancels the next, leaving W = kqq₀(1/r − 1/∞) = kqq₀/r.
6. 6★
   So, V = W/q₀ = (1/4πε₀) q/r. The potential falls as 1/r and has the same sign as q.

★ key step

Derivation: potential due to a short dipole

−q is at A and +q at B, with AB = 2l and centre O. P is far away, with OP = r and angle BOP = θ.

1. 1
   Let AP = r₁ (from −q) and BP = r₂ (from +q). Potentials add as numbers: V = kq/r₂ − kq/r₁ = kq(1/r₂ − 1/r₁).
2. 2
   Drop perpendiculars from A and B onto OP. For a small dipole, r₁ ≈ r + l cos θ and r₂ ≈ r − l cos θ.
3. 3
   So 1/r₂ − 1/r₁ = (r₁ − r₂)/(r₁r₂) = 2l cos θ/(r² − l² cos²θ).
4. 4
   For r much larger than l, drop l² cos²θ. Then V = kq × 2l cos θ/r².
5. 5★
   With p = 2lq: V = (1/4πε₀) p cos θ/r². It falls as 1/r², faster than the 1/r of a point charge.
6. 6
   Special cases: on the axis on the +q side (θ = 0°), V = +p/(4πε₀r²); on the −q side (θ = 180°), V = −p/(4πε₀r²); on the equatorial line (θ = 90°), V = 0.

★ key step

How to draw the dipole potential figure

• Draw AB horizontal, with −q at A, +q at B and the centre O. Mark AB = 2l and the arrow p from A to B.
• Draw a long line from O up to the right, at an angle θ above OB. Mark P on it and label OP = r.
• Join AP and BP, and label them r₁ and r₂.
• Drop perpendiculars from A and B onto the line OP, extended if needed. Label each short piece of OP l cos θ.

Worked example: potential and work

Worked example: potential of a short dipole