3cElectric Potential and Voltage
Explore electric potential energy, the scalar concept of voltage, equipotential surfaces, and how capacitors store electrical energy.
Voltage is the concept that makes electrical engineering possible. Every battery, power supply, and circuit element is characterized by its voltage. Capacitors store energy in electric fields and are essential in cameras (flash), defibrillators, and computer memory. Understanding potential is the bridge between electrostatics and circuit analysis.
Why does a 9-volt battery have "9 volts" — and what does that actually mean for the energy of charges moving through a circuit?
Lesson Overview
Electric potential energy (U) is the energy stored in a system of charges due to their positions. Electric potential (V) is the electric potential energy per unit charge: V = U/q. The SI unit is the volt (1 V = 1 J/C). The potential due to a point charge Q at distance r is V = kQ/r. Potential difference (voltage) between two points is ΔV = V_B − V_A = −W_{ A→B}/q. Equipotential surfaces are surfaces of constant potential — no work is done moving a charge along an equipotential. They are always perpendicular to electric field lines. A capacitor stores charge and energy; its capacitance is C = Q/V (farads), and the energy stored is U = ½CV².
Key Equations
Worked Examples
Find the electric potential at a point 0.30 m from a +5.0 μC charge.
A charge of +2.0 μC is moved from point A (V = 100 V) to point B (V = 400 V). Find the work done by the electric field and the work done by an external agent.
Two charges: q₁ = +3.0 μC at x = 0 and q₂ = −3.0 μC at x = 0.40 m. Find the electric potential at the midpoint (x = 0.20 m).
A parallel-plate capacitor has plates separated by 5.0 mm and a voltage of 200 V across it. Find the electric field between the plates.
A capacitor with capacitance 10 μF is charged to 12 V. Find (a) the charge stored and (b) the energy stored.
Guided Problems
Find the electric potential at a point 0.50 m from a −4.0 μC charge. Is the potential positive or negative?
Hint: V = kQ/r. For a negative charge, Q is negative, so V is negative. Potential is a scalar — just substitute the signed value of Q.
An electron (q = −e) is accelerated through a potential difference of 1000 V. Find the kinetic energy gained by the electron.
Hint: Work done by field = qΔV. For an electron moving from low to high potential (ΔV = +1000 V), W = qΔV = (−e)(+1000) = −1.6×10⁻¹⁶ J... wait, think about direction. The electron accelerates from high to low potential (ΔV = −1000 V for the electron's path). Use ΔKE = |q|ΔV = eV.
A capacitor stores 50 μC of charge when connected to a 25 V battery. Find its capacitance and the energy stored.
Hint: C = Q/V. Then U = ½CV² or U = Q²/(2C) or U = ½QV — all three give the same answer.
The electric field between two parallel plates is 6.0×10³ N/C and the plates are 8.0 mm apart. Find the potential difference between the plates.
Hint: For a uniform field: E = ΔV/d → ΔV = Ed.
Explain why no work is done when a charge moves along an equipotential surface.
Hint: Work = qΔV. On an equipotential, ΔV = 0, so W = 0 regardless of the path or the charge.
Key Vocabulary
Electric Potential (V)
The electric potential energy per unit charge at a point: V = U/q. Unit: volt (V = J/C). It is a scalar quantity. The potential due to a point charge Q at distance r is V = kQ/r.
Example: Near a +1 μC charge at 0.10 m, V = (9×10⁹)(1×10⁻⁶)/0.10 = 90,000 V = 90 kV.
Potential Difference (Voltage)
The difference in electric potential between two points: ΔV = V_B − V_A. Measured in volts. Represents the work done per unit charge moving between the points.
Example: A 9 V battery maintains a potential difference of 9 V between its terminals — 9 joules of work per coulomb of charge.
Equipotential Surface
A surface on which the electric potential is the same at every point. No work is done moving a charge along an equipotential. Equipotentials are always perpendicular to electric field lines.
Example: For a point charge, equipotential surfaces are concentric spheres. For a uniform field, they are parallel planes perpendicular to the field.
Electron Volt (eV)
A unit of energy equal to the kinetic energy gained by one electron accelerated through 1 volt: 1 eV = 1.6×10⁻¹⁹ J. Commonly used in atomic and nuclear physics.
Example: An electron accelerated through 1000 V gains 1000 eV = 1.6×10⁻¹⁶ J of kinetic energy.
Capacitance
The ability of a device to store charge per unit voltage: C = Q/V. Unit: farad (F = C/V). For a parallel-plate capacitor: C = ε₀A/d.
Example: A 100 μF capacitor charged to 5 V stores Q = CV = 500 μC of charge.
Dielectric
An insulating material placed between capacitor plates that increases capacitance by a factor κ (the dielectric constant): C = κε₀A/d. Also increases the maximum voltage the capacitor can withstand.
Example: A capacitor with a paper dielectric (κ ≈ 3.5) has 3.5 times the capacitance of the same capacitor in vacuum.
Workbook Check — Interactive Quiz
Interactive Practice — 5 Questions
The electric potential due to a point charge Q at distance r is:
A +3.0 μC charge is moved from V = 200 V to V = 500 V. The work done by the electric field is:
Equipotential surfaces are always:
A 20 μF capacitor is charged to 10 V. The energy stored is:
The relationship between electric field and potential in a uniform field is:
Independent Practice
Find the electric potential at a point 0.40 m from a +6.0 μC charge and at a point 0.40 m from a −6.0 μC charge.
A proton is accelerated from rest through a potential difference of 5000 V. Find its final kinetic energy and speed. (m_p = 1.67×10⁻²⁷ kg)
A parallel-plate capacitor has plate area 0.020 m² and separation 2.0 mm. Find its capacitance (ε₀ = 8.85×10⁻¹² F/m). If charged to 100 V, find the charge and energy stored.
Explain why the electric field inside a conductor is zero and the surface of a conductor is an equipotential surface.
★ Two capacitors C₁ = 4.0 μF and C₂ = 6.0 μF are connected (a) in series and (b) in parallel to a 12 V battery. Find the equivalent capacitance, total charge, and energy stored in each case.
ChallengeCommon Mistakes
Treating electric potential as a vector and adding it with direction
Electric potential is a scalar. Add potentials from multiple charges algebraically (with signs): V_total = V₁ + V₂ + V₃ + …
Confusing V = kQ/r (potential, 1/r) with E = kQ/r² (field, 1/r²)
Potential falls off as 1/r; field falls off as 1/r². At double the distance, V halves but E decreases by factor 4.
Thinking work done by the field is W = qV (not qΔV)
Work done by the field moving charge q from A to B is W = q(V_A − V_B) = qΔV. Only potential DIFFERENCES are physically meaningful.
Forgetting the sign of the charge when calculating potential or work
V = kQ/r uses the signed value of Q. W = qΔV uses the signed value of q. A negative charge moving to higher potential gains kinetic energy.
Math Tips
V = kQ/r (scalar, 1/r). E = kQ/r² (vector, 1/r²). Potential is easier to work with — it is a scalar sum.
Work by field: W = q(V_A − V_B). Work by external agent: W_ext = q(V_B − V_A) = −W_field.
Energy of a charge: ΔKE = qΔV (for a charge moving through potential difference ΔV). For an electron: ΔKE = eV (in eV or joules).
Capacitor formulas: C = Q/V; U = ½CV² = ½QV = Q²/(2C). All three energy formulas are equivalent.