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Electric potential is one of the most powerful concepts in electrostatics. While electric field strength tells you about the force a charge would experience, electric potential tells you about the energy landscape — how much work would be done in moving a charge from one place to another. Because potential is a scalar quantity, calculations involving multiple charges are often simpler when you work with potential rather than field.
The electric potential V at a point in a field is defined as the work done per unit positive charge in bringing a small test charge from infinity to that point.
For a point charge Q, the electric potential at a distance r is:
V = kQ / r = Q / (4πε₀r)
where k = 8.99 × 10⁹ N m² C⁻².
Because potential is a scalar, finding the total potential due to multiple charges is simpler than finding the total field — you just add the values algebraically (no vector components needed).
The potential V = kQ/r follows a 1/r relationship, whereas the field E = kQ/r² follows a 1/r² relationship. This means:
The potential difference between two points is the work done per unit charge in moving a charge between those points:
ΔV = W / Q
where W is the work done (J) and Q is the charge (C).
If a charge Q moves through a potential difference ΔV, the work done on it is:
W = QΔV
This is the basis for the electronvolt (eV): the energy gained by an electron accelerated through a potential difference of 1 V.
1 eV = 1.6 × 10⁻¹⁹ J
An electron is accelerated from rest through a potential difference of 5000 V. What speed does it reach? (mₑ = 9.11 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C)
Work done on the electron: W = QΔV = 1.6 × 10⁻¹⁹ × 5000 = 8.0 × 10⁻¹⁶ J
This equals the kinetic energy gained: ½mv² = 8.0 × 10⁻¹⁶
v² = 2 × 8.0 × 10⁻¹⁶ / 9.11 × 10⁻³¹ = 1.76 × 10¹⁵
v = 4.2 × 10⁷ m s⁻¹
This is about 14% of the speed of light — at these speeds, relativistic effects start to become noticeable.
An alpha particle (α) with charge +2e and kinetic energy 5.0 MeV is fired directly at a gold nucleus (charge +79e). Find the distance of closest approach.
At closest approach, all kinetic energy has been converted to electric potential energy:
E_k = kQ₁Q₂ / r
r = kQ₁Q₂ / E_k
r = (8.99 × 10⁹ × 2 × 1.6 × 10⁻¹⁹ × 79 × 1.6 × 10⁻¹⁹) / (5.0 × 10⁶ × 1.6 × 10⁻¹⁹)
r = (8.99 × 10⁹ × 4.045 × 10⁻³⁶) / (8.0 × 10⁻¹³)
r = 3.636 × 10⁻²⁶ / 8.0 × 10⁻¹³ = 4.5 × 10⁻¹⁴ m
This is about 45 fm — much larger than the gold nucleus radius (~7 fm), confirming the alpha particle does not penetrate the nucleus at this energy.
An equipotential surface (or line, in 2D) is a surface on which every point has the same electric potential. Key properties:
The spacing of equipotential lines tells you about the field strength — closer equipotentials mean a stronger field (the potential is changing more rapidly with distance).
Think of equipotentials like contour lines on a map. A steep hill has closely spaced contour lines (large change in height over a small horizontal distance). Similarly, closely spaced equipotentials indicate a strong electric field (large change in potential over a small distance).
A charge placed on an equipotential “hill” will “roll downhill” towards lower potential (for positive charges) or “uphill” towards higher potential (for negative charges).
The electric field strength is related to how rapidly the potential changes with distance:
E = −dV / dr
The negative sign indicates that the field points in the direction of decreasing potential — from high potential to low potential.
For a uniform field between parallel plates:
E = V / d
which is consistent with E = −dV/dr since the potential drops linearly from one plate to the other.
For a point charge, we can verify: V = kQ/r, so dV/dr = −kQ/r², and E = −(−kQ/r²) = kQ/r², which matches the field of a point charge.
The electric potential energy of a system of two point charges Q₁ and Q₂ separated by a distance r is:
Eₙ = kQ₁Q₂ / r = Q₁Q₂ / (4πε₀r)
This is the work done in bringing the charges from infinity to the separation r.
Calculate the electric potential energy of a proton and electron separated by 5.3 × 10⁻¹¹ m (the Bohr radius).
Eₙ = kQ₁Q₂ / r = (8.99 × 10⁹ × (+1.6 × 10⁻¹⁹) × (−1.6 × 10⁻¹⁹)) / (5.3 × 10⁻¹¹)
Eₙ = (8.99 × 10⁹ × (−2.56 × 10⁻³⁸)) / (5.3 × 10⁻¹¹)
Eₙ = −2.30 × 10⁻²⁸ / 5.3 × 10⁻¹¹ = −4.3 × 10⁻¹⁸ J
The negative value makes physical sense — you would have to do work (add energy) to separate the proton and electron.
Converting to electronvolts: Eₙ = −4.3 × 10⁻¹⁸ / 1.6 × 10⁻¹⁹ = −27 eV
This is the ionisation energy of hydrogen — the energy needed to completely remove the electron from the atom.
The work done in moving a charge Q from a point at potential V₁ to a point at potential V₂ is:
W = Q(V₂ − V₁) = QΔV
If the charge moves to a higher potential, work is done on it. If it moves to a lower potential, it releases energy (which becomes kinetic energy if the charge is free to accelerate).
A proton is moved from a point where the electric potential is +200 V to a point where the potential is +500 V. Calculate the work done and state whether the work is done on or against the proton.
W = QΔV = 1.6 × 10⁻¹⁹ × (500 − 200) = 1.6 × 10⁻¹⁹ × 300 = 4.8 × 10⁻¹⁷ J
The proton moves to a higher potential. Since positive charges naturally move from high to low potential, work must be done on the proton (by an external agent) to move it to the higher potential.
Three charges are arranged at the corners of an equilateral triangle with side 0.10 m: Q₁ = +2.0 nC, Q₂ = +2.0 nC, Q₃ = −4.0 nC. Find the total electric potential at the centre of the triangle.
The distance from each corner to the centre of an equilateral triangle with side a is r = a / √3 = 0.10 / √3 = 0.0577 m.
V_total = k(Q₁/r + Q₂/r + Q₃/r) = k(Q₁ + Q₂ + Q₃) / r
V_total = (8.99 × 10⁹ × (2.0 + 2.0 − 4.0) × 10⁻⁹) / 0.0577
V_total = (8.99 × 10⁹ × 0) / 0.0577 = 0 V
The potential is zero because the charges sum to zero. However, the electric field at this point is not zero — the field contributions do not cancel because they are vectors with different directions.
For two like charges, the potential energy Eₙ = kQ₁Q₂/r is positive and increases as r decreases (it gets harder to push them together). The graph of Eₙ against r is a positive curve that approaches zero from above as r → ∞.
For two opposite charges, Eₙ is negative and becomes more negative as r decreases. The graph is a negative curve that approaches zero from below as r → ∞. The system is bound — you must add energy to separate the charges to infinity.
Edexcel 9PH0 specification Topic 7 — Electric and Magnetic Fields covers electric potential as the work done per unit positive charge in bringing a small test charge from infinity to a point in the field; the radial-field result V=kQ/r for a point charge; the pair potential energy U=kQ1Q2/r; equipotential surfaces and their perpendicularity to field lines; and the differential link E=−dV/dr (refer to the official Pearson Edexcel specification document for exact wording). The topic is examined predominantly in 9PH0 Paper 2 (Advanced Physics II), with synoptic reach into Paper 3 through equipotential-mapping practicals and into Paper 1 wherever the electronvolt is used as an energy unit. The Edexcel formula booklet lists V=kQ/r, E=−dV/dr (and E=V/d) and the Coulomb constant k=1/(4πε0), but does not list U=kQ1Q2/r as a separate equation — candidates derive it from W=QV.
Question (8 marks):
(a) A point charge Q1=+6.0×10−9 C is fixed at the origin in vacuum. Calculate the electric potential at point P, a distance 0.20 m from the charge, and state whether work would be done on or by an external agent in moving a small +1.5×10−9 C test charge from infinity to P. (4)
(b) A second fixed point charge Q2=−4.0×10−9 C is now placed 0.30 m from Q1 along the same line, with P sitting between them at 0.20 m from Q1 and 0.10 m from Q2. Calculate the new total potential at P and the work done in moving the +1.5×10−9 C test charge from infinity to P. (4)
Solution with mark scheme:
(a) Step 1 — apply V=kQ/r.
V=rkQ1=0.20(8.99×109)(6.0×10−9)
M1 — correct substitution into the radial-potential formula. The source charge Q1 goes in the numerator; the test charge does not appear here.
Step 2 — evaluate.
V=0.2053.94≈270 V
A1 — magnitude correct to two significant figures with units stated.
Step 3 — interpret the sign.
Because Q1 is positive, V at P is positive (+270 V). Bringing a positive test charge from infinity (where V=0) to a point of higher positive potential requires work to be done against the repulsive field. The work is therefore done on the test charge by an external agent.
B1 — correct identification that work is done on the test charge, justified by the like-sign repulsion.
B1 — explicit reference to the convention V∞=0 as the reason the sign of V at P matters.
(b) Step 1 — calculate V2 at P due to Q2.
V2=r2kQ2=0.10(8.99×109)(−4.0×10−9)≈−360 V
M1 — correct substitution including the negative sign on Q2. Potential is scalar, so the sign is carried through — the central distinction from field-strength calculations.
Step 2 — sum scalar potentials.
Vtotal=V1+V2=270+(−360)=−90 V
A1 — algebraic addition without vector resolution. Many candidates instinctively try to "add components" — there are none, because V is scalar.
Step 3 — work done on the test charge.
W=qVtotal=(1.5×10−9)(−90)=−1.35×10−7 J≈−1.4×10−7 J
M1 — correct use of W=qV with the test charge here.
A1 — the negative sign means the field does work on the test charge; an external agent would do negative work (allow it to fall in). A clean direction statement earns the final A1.
Total: 8 marks (M3 A3 B2).
Question (6 marks): Two horizontal parallel plates are separated by d=12 mm with a potential difference of 240 V between them, the upper plate being positive. A small dust particle of mass 4.0×10−15 kg carrying charge −2e is released from rest just below the upper plate.
(a) Sketch (in words) the equipotential surfaces between the plates and state how the potential varies with vertical position y measured downward from the upper plate. (2)
(b) Calculate the work done by the electric field on the particle as it travels from the upper plate to the lower plate, and hence its kinetic energy on arrival. (4)
Mark scheme decomposition by AO:
(a)
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