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Unit VII · Electric potential and capacitors (lesson 16)
10 min readConcept sheetSection A objective parts (true or false, fill in the blank); Section B very short answer (2 marks) on equipotential surfaces, E = −ΔV/Δr or electrostatic shielding

Field and potential, equipotential surfaces, and conductors

The field is the negative potential gradient, E = −ΔV/Δr, and crosses equipotential surfaces at right angles. Inside a conductor, E = 0.

Field and potentialE = −ΔV/Δr; in a uniform field E = (V_A − V_B)/d; unit V m⁻¹, the same as N C⁻¹
Equipotential surfaceEvery point on it has the same potential, so no work is done moving a charge along it
ShapesPoint charge: concentric spheres. Uniform field: parallel planes at right angles to E
Conductor in electrostaticsE = 0 inside; any charge sits on the surface; E just outside is at right angles to the surface

Derivation: relation between E and V

A and B are two close points in a uniform field E, a small distance Δr apart, with B farther along the field.

  1. 1
    Move a unit positive charge from A to B. The field pushes it along the motion, so the outside agent must push against the motion.
  2. 2
    Work done by the agent on the unit charge = EΔr cos 180° = −EΔr. This is the potential difference ΔV = V(B) − V(A).
  3. 3★
    So, E = −ΔV/Δr. The field is the negative potential gradient: it points the way the potential falls fastest.
  4. 4
    Between two points a distance d apart along a uniform field: E = (V_A − V_B)/d.
  5. 5
    Hence the unit of E can also be written V m⁻¹, the same as N C⁻¹. Where V is the same everywhere, E = 0.

★ key step

Equipotential surfaces

An equipotential surface is a surface on which every point has the same potential. For a point charge, V = kq/r depends only on r, so its equipotentials are concentric spheres. In a uniform field they are parallel planes at right angles to the field. For a dipole, the whole equatorial plane is the V = 0 surface. Moving a charge along an equipotential needs no work, because W = q × ΔV = 0. So the field has no part along the surface: E is always at right angles to an equipotential surface.

Properties of equipotential surfaces

  • No work is done in moving a charge along an equipotential surface.
  • The electric field is always perpendicular to the surface at every point.
  • Two equipotential surfaces never intersect. At the crossing, one point would have two potentials.
  • For equal steps of potential, the surfaces are closer together where the field is strong.
  • The surface of a conductor in electrostatics is an equipotential surface.

How to draw equipotential surfaces

  1. 1
    Point charge: dashed concentric circles round +q. Draw solid radial field lines crossing them at right angles. For equal steps of V, the circles get farther apart going outward.
  2. 2
    Uniform field: solid horizontal field lines with arrows; dashed vertical straight lines, equally spaced, crossing them at right angles.
  3. 3
    Dipole: dashed closed curves round each charge, crowded near the charges. The perpendicular bisector is the V = 0 surface. Field lines cross every dashed curve at right angles.
  4. 4
    Two equal positive charges: dashed closed curves round each charge. Farther out they merge into one curve shaped like a peanut, and far away they look like circles.

Conductors in an electric field

  • A conductor, such as a metal, has free electrons that move easily. An insulator, such as wood, glass, mica or ebonite, has none.
  • In an outside field E, the free electrons drift against E and pile up on one face. The opposite face is left positive.
  • These surface charges make a field opposite to E. The electrons stop when the two fields cancel, in about 10⁻¹⁶ s.
  • So, in electrostatic equilibrium, E = 0 inside a conductor, and the whole conductor is at one potential.
  • Any charge on a conductor sits on its outer surface, and the field just outside is at right angles to the surface.
  • Electrostatic shielding: the field inside an empty cavity in a conductor is zero, whatever the field outside. So a car or bus is safer than open ground in a lightning storm, and delicate instruments are kept in metal boxes.

Worked example: field from potentials

Problem

(a) Two large parallel plates 5 mm apart are at 30 V and 10 V. Find the field between them. (b) How much work does the field do on a charge of +2 μC that moves from the 30 V plate to the 10 V plate?

Working
  1. (a) The field is uniform: E = (V_A − V_B)/d = (30 − 10) V ÷ 0.005 m = 4000 V m⁻¹.
  2. It points from the 30 V plate towards the 10 V plate, the way the potential falls.
  3. (b) Work done by the field = q[V(A) − V(B)] = 2 × 10⁻⁶ C × 20 V = 4 × 10⁻⁵ J.
  4. The positive charge moves from high to low potential, so the field does positive work on it.
Answer

(a) 4 × 10³ V m⁻¹, from the 30 V plate towards the 10 V plate; (b) 4 × 10⁻⁵ J

In real life

In a thunderstorm, sitting inside a closed car or bus is far safer than standing in an open field, because its metal body shields the inside. Cable TV wires and sensitive electronics use metal braids and casings for the same reason.

Exam angle

1 marks: “How much work is done in moving a charge of 5 μC through 10 cm along an equipotential surface?” Zero, because there is no potential difference between any two points of an equipotential surface.

2 marks: “Why can two equipotential surfaces never intersect?” At a point of intersection the potential would have two values, and E, which is normal to each surface, would have two directions. Both are impossible.

2 marks: “What is electrostatic shielding? Give one use of it.” Shielding is protecting a region from outside electric fields by surrounding it with a conductor: the field inside the cavity is zero. Use: delicate instruments are kept in metal cases; a car protects its passengers during lightning.

Remember it

Field runs downhill, at right angles to the contours: like water on a hill map, E flows from high V to low V, crossing equipotentials at 90°.