Electric potential, potential difference and potential due to a point charge
Electric potential at a point is the work done per unit positive charge to bring a charge there from infinity. For a point charge, V = q/(4πε₀r).
Why we use potential
The electrostatic force is conservative. So the work done in moving a charge depends only on the start and end points, not on the path. This work is stored as potential energy. The potential difference between two points is the work done per unit positive charge to move a charge from one point to the other. An external force does this work without accelerating the charge. The electric potential at a point is this work when the charge is brought from infinity, where the potential is taken as zero. 1 volt is the potential at a point when 1 J of work brings 1 C of charge there from infinity.
Derivation: potential due to a point charge
A point charge Q sits at the origin O. Find the potential at a point P at distance r.
- 1Bring a unit positive test charge from infinity to P along the line OP, so slowly that it does not accelerate.
- 2At a distance r′ from O, the electric force on the unit charge is Q/(4πε₀r′²), pointing away from O (for Q > 0).
- 3The external force must be equal and opposite. Its work from infinity to r is W = −∫ Q/(4πε₀r′²) dr′, with r′ going from ∞ to r.
- 4Since ∫ dr′/r′² = −1/r′, W = (Q/4πε₀)[1/r′] taken from ∞ to r = Q/(4πε₀r) − 0.
- 5★So, V = (1/4πε₀) Q/r. The formula holds for positive and negative Q.
★ key step
Electric field vs electric potential
They are linked by E = −dV/dr: the field points the way the potential falls fastest.
Worked example: potential and work
(a) Find the potential at a point 30 cm from a charge of +6 nC in air. (b) How much work is needed to bring a charge of +2 μC from infinity to that point?
- (a) V = kq/r = 9 × 10⁹ × 6 × 10⁻⁹ ÷ 0.30.
- V = 54 ÷ 0.30 = 180 V.
- (b) W = qV = 2 × 10⁻⁶ C × 180 V = 3.6 × 10⁻⁴ J.
(a) 180 V; (b) 3.6 × 10⁻⁴ J
Worked example: potential difference
Points A and B are 10 cm and 20 cm from a charge of +2 nC. Find V(A) − V(B), and the work needed to move a +3 μC charge from B to A.
- V(A) = 9 × 10⁹ × 2 × 10⁻⁹ ÷ 0.10 = 18 ÷ 0.10 = 180 V.
- V(B) = 18 ÷ 0.20 = 90 V.
- V(A) − V(B) = 180 − 90 = 90 V.
- Work from B to A: W = q[V(A) − V(B)] = 3 × 10⁻⁶ × 90 = 2.7 × 10⁻⁴ J.
V(A) − V(B) = 90 V; work = 2.7 × 10⁻⁴ J
How to draw V and E against r for a point charge
- Put r on the horizontal axis. Draw the V curve and the E curve on separate axes, or label them clearly.
- V curve: falls as 1/r, so it drops gently.
- E curve: falls as 1/r², so it drops more steeply. Doubling r halves V but quarters E.
- Both curves come closer and closer to the r-axis without touching it.
A 9 V battery in a smoke alarm gives every coulomb of charge 9 J of energy between its terminals. Birds sit safely on a single live wire because both their feet are at the same potential, so no current passes through them.
1 marks: “Is electric potential a scalar or a vector? Give its SI unit.” A scalar. SI unit volt (1 V = 1 J C⁻¹).
2 marks: “Define electric potential at a point. How much work is done in bringing a 2 μC charge from infinity to a point at a potential of 50 V?” The work done per unit positive charge to bring it from infinity without acceleration. W = qV = 2 × 10⁻⁶ × 50 = 1 × 10⁻⁴ J.
3 marks: “Derive an expression for the electric potential at a distance r from a point charge q.” Bring a unit charge from ∞ to r: W = −∫ q/(4πε₀r′²) dr′ from ∞ to r = q/(4πε₀r). So V = q/(4πε₀r).
Field is newtons per coulomb, potential is joules per coulomb. Point charge: V = kq/r (one r), E = kq/r² (r squared).