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Electric potential provides an energy-based description of electric fields, just as gravitational potential does for gravitational fields. This topic (AQA 3.7.3) introduces the concept of potential, equipotentials, and energy changes when charges move through fields, culminating in Millikan's oil drop experiment.
Key Definition: The electric potential (V) at a point in an electric field is the work done per unit positive charge in bringing a small positive test charge from infinity to that point.
For a point charge Q:
V = kQ/r = Q/(4πε₀r)
where V is the electric potential (V or J C⁻¹), k = 8.99 × 10⁹ N m² C⁻², Q is the source charge (C), and r is the distance from the charge (m).
Key differences from gravitational potential:
Question: Calculate the electric potential at a distance of 0.20 m from a point charge of +4.0 μC.
Solution:
V = kQ/r = (8.99 × 10⁹ × 4.0 × 10⁻⁶) / 0.20
V = 3.596 × 10⁴ / 0.20
V = 1.80 × 10⁵ V = 180 kV
Since the charge is positive, the potential is positive. A positive test charge would need to have work done on it to bring it from infinity to this point (against the repulsive force).
Equipotential surfaces connect points of equal electric potential.
Described diagram — Equipotentials around a positive charge: Concentric circles centred on the charge, with potential values decreasing (but remaining positive) further from the charge. Field lines radiate outward and are perpendicular to every equipotential circle.
Described diagram — Equipotentials between parallel plates: Equally spaced straight lines parallel to the plates. The potential decreases uniformly from the positive plate to the negative plate. Field lines are perpendicular to these equipotential lines, running from positive to negative plate.
Properties:
Since electric potential is a scalar, the total potential at a point due to multiple charges is simply the algebraic sum:
V_total = V₁ + V₂ + V₃ + ... = Σ kQᵢ/rᵢ
This is much simpler than adding fields (which requires vector addition).
Question: A charge of +6.0 μC is placed at the origin and a charge of −3.0 μC is placed 0.40 m away. Calculate the electric potential at the midpoint between the charges.
Solution:
At the midpoint, the distance from each charge is 0.20 m.
V₁ = kQ₁/r₁ = (8.99 × 10⁹ × 6.0 × 10⁻⁶) / 0.20 = 2.70 × 10⁵ V
V₂ = kQ₂/r₂ = (8.99 × 10⁹ × (−3.0 × 10⁻⁶)) / 0.20 = −1.35 × 10⁵ V
V_total = V₁ + V₂ = 2.70 × 10⁵ + (−1.35 × 10⁵) = 1.35 × 10⁵ V = 135 kV
Note: Although the potential at the midpoint is not zero, the field at the midpoint is not zero either (both fields point in the same direction — away from the positive charge and towards the negative charge).
The electric field strength is the negative of the potential gradient:
E = −dV/dr
In a uniform field (parallel plates):
E = V/d (magnitude)
On a V-r graph for a point charge, the gradient at any point gives −E at that distance.
The electric potential energy of a charge q at a point where the potential is V:
E_p = qV
For two point charges Q₁ and Q₂ separated by distance r:
E_p = kQ₁Q₂/r = Q₁Q₂/(4πε₀r)
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