What this chapter is about
This chapter builds on your understanding of electric charges and fields by introducing two related quantities: electrostatic potential and capacitance. While the previous chapter focused on the electric field as a vector quantity, this chapter presents electric potential as a scalar quantity that simplifies many calculations. You learn how work done in moving charges relates to potential energy and potential difference.
The second major theme is capacitance — the ability of a system to store electric charge and energy. You study how capacitors work, how different arrangements of conductors affect capacitance, and how dielectric materials placed between capacitor plates change their behaviour. These ideas connect directly to practical devices like camera flashes, defibrillators and electronic circuits.
After studying this chapter, you should be able to calculate potential at a point due to various charge configurations, relate potential energy to work done, analyse capacitor combinations, and understand why dielectrics are used in real capacitors.
Key ideas
- Electric potential at a point is the work done per unit positive charge in bringing a small test charge from infinity to that point. It is a scalar quantity measured in volts (V), where 1 V = 1 J/C.
- Potential difference between two points equals the work done per unit charge in moving a test charge between those points. Current flows from higher to lower potential.
- Equipotential surfaces are surfaces where every point has the same potential. The electric field is always perpendicular to an equipotential surface, and no work is done when a charge moves along such a surface.
- The potential due to a point charge Q at distance r is V = kQ/r = Q/(4πε₀r), where k = 9 × 10⁹ N m² C⁻².
- Capacitance C of a conductor is defined as C = Q/V, the ratio of charge stored to potential. The SI unit is farad (F), where 1 F = 1 C/V.
- For a parallel plate capacitor with plate area A and separation d, the capacitance is C = ε₀A/d in vacuum, and C = Kε₀A/d when a dielectric of constant K fills the gap.
- Energy stored in a charged capacitor is U = (1/2)CV² = (1/2)QV = Q²/(2C).
- Capacitors in parallel add directly: C_total = C₁ + C₂ + ... . Capacitors in series combine as 1/C_total = 1/C₁ + 1/C₂ + ... .
Formulas and facts to remember
Potential due to a point charge: V = Q/(4πε₀r) — potential decreases as 1/r from a point charge.
Potential energy of two charges: U = Q₁Q₂/(4πε₀r) — the work needed to assemble the configuration.
Relation between field and potential: E = −dV/dr along the direction of the field — field points from high to low potential.
Capacitance definition: C = Q/V — larger capacitance means more charge stored per volt.
Parallel plate capacitor: C = ε₀A/d — capacitance increases with area and decreases with separation.
Effect of dielectric: C becomes KC₀, where K is the dielectric constant and C₀ is capacitance without the dielectric.
Energy stored: U = (1/2)CV² — this energy resides in the electric field between the plates.
Series combination: 1/C_eq = 1/C₁ + 1/C₂ — equivalent capacitance is smaller than the smallest individual capacitor.
Parallel combination: C_eq = C₁ + C₂ — equivalent capacitance is the sum of individual capacitances.
Worked examples
Example 1: Potential due to two charges
Two point charges, +4 μC and −2 μC, are placed 20 cm apart. Find the potential at the midpoint.
Solution: Distance from each charge to midpoint = 10 cm = 0.10 m.
Potential due to +4 μC: V₁ = kQ₁/r = (9 × 10⁹)(4 × 10⁻⁶)/0.10 = 3.6 × 10⁵ V.
Potential due to −2 μC: V₂ = kQ₂/r = (9 × 10⁹)(−2 × 10⁻⁶)/0.10 = −1.8 × 10⁵ V.
Total potential at midpoint: V = V₁ + V₂ = 3.6 × 10⁵ − 1.8 × 10⁵ = 1.8 × 10⁵ V.
Since potential is a scalar, we add the contributions algebraically.
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Example 2: Capacitors in combination
Three capacitors of 2 μF, 3 μF and 6 μF are connected in series across a 12 V battery. Find the equivalent capacitance and charge on each capacitor.
Solution: For series combination: 1/C_eq = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1.
So C_eq = 1 μF.
Total charge: Q = C_eq × V = 1 × 10⁻⁶ × 12 = 12 μC.
In series, the same charge flows through each capacitor. Therefore, each capacitor carries 12 μC.
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Example 3: Energy stored with a dielectric
A parallel plate capacitor of capacitance 50 pF is charged to 100 V and then disconnected. A dielectric slab of K = 5 is inserted fully between the plates. Find the new potential and energy stored.
Solution: Initial charge: Q = CV = 50 × 10⁻¹² × 100 = 5 × 10⁻⁹ C.
Since the capacitor is disconnected, charge remains constant at 5 nC.
New capacitance: C' = KC = 5 × 50 = 250 pF.
New potential: V' = Q/C' = 5 × 10⁻⁹ / 250 × 10⁻¹² = 20 V.
Initial energy: U = (1/2)CV² = (1/2)(50 × 10⁻¹²)(100)² = 2.5 × 10⁻⁷ J.
Final energy: U' = (1/2)C'V'² = (1/2)(250 × 10⁻¹²)(20)² = 5 × 10⁻⁸ J.
Energy decreases because work is done by the electric field in pulling the dielectric inward.
Common mistakes
Treating potential as a vector → Potential is a scalar; add contributions with their signs, not as vectors.
Using the series formula for parallel capacitors → Remember: series uses reciprocals, parallel uses direct addition.
Forgetting the sign of potential for negative charges → V = kQ/r includes the sign of Q; negative charges give negative potential.
Assuming charge changes when a disconnected capacitor gets a dielectric → Charge stays constant when disconnected; potential and energy change instead.
Ignoring units when calculating capacitance → Keep distances in metres and charges in coulombs to get farads directly.
Quick revision
- Electric potential is work per unit charge; it is a scalar measured in volts.
- Field points from high potential to low potential and is perpendicular to equipotential surfaces.
- Capacitance C = Q/V; a dielectric increases capacitance by factor K.
- Series capacitors: same charge, voltages add; parallel capacitors: same voltage, charges add.
- Energy stored in a capacitor is (1/2)CV², residing in the electric field.