Coulomb's Law

Coulomb's Law

• For charges at rest, Coulomb's Law was formulated and experimentally defined in 1784 by French physicist Charles Augustin de Coulomb
• Assumes two point charges (Q1 and Q2) are in free space (a vacuum), separated by distance r
• Electric force observations between charges at rest:
  • Like charges repel; unlike charges attract
  • Force acts along the line between charges
  • Force magnitude is inversely proportional to the square of the distance between the two charges
  • Force magnitude is proportional to the product of the magnitudes of the charges
• The electric force formula is 𝐹 ∝ (𝑄1 𝑄2)/𝑟², measured in Newtons (N)
• Charge polarity determines if the force is attractive or repulsive
  • If Q1 and Q2 have the same polarity, the force is repulsive and positive
  • If Q1 and Q2 have opposite polarities, the force is attractive and negative
• The unit of charge is the coulomb (C), experimentally defined from electric current where 1 amp = 1 coulomb flowing for 1 second (1 A = 1 C/s)
• The charge on an electron is 1.602 × 10−19 C, therefore 1 C is the charge of 6.24 × 10^18 electrons
• In free space (a vacuum), a point charge's force acts equally in all directions in three dimensions, distributing over a spherical surface, where sphere surface area A = 4πr²
• Material properties affect the force, so a fully-defined law includes them
• The fully defined law: 𝐹⃗ = 𝑘𝑒 (𝑄1 𝑄2)/𝑟² = (𝑄1 𝑄2)/(4𝜋𝑟²𝜀0 𝜀𝑟)
  • 𝜀0 is free space permittivity (vacuum), equals 8.854 × 10−12 C² ⁄Nm²
  • 𝜀𝑟 is the surrounding material’s relative permittivity (dimensionless); 𝜀𝑟 = 1 for free space
• For two isolated point charges, Coulomb’s Law is 𝐹⃗ = (𝑄1 𝑄2)/(4𝜋𝑟²𝜀0 𝜀𝑟) (N)
• Force is a vector quantity, having both magnitude and direction

Forces Acting on Multiple Charges (Superposition)

• With multiple charges, any given charge feels a force that is the vector sum of forces on that charge from other charges
• This is the principle of superposition
• Magnitudes alone cannot be summed unless all forces act along the same line because forces are vector quantities

Electric Field

• For two point charges (𝑄1 and 𝑄2 ) in free space (𝜀𝑟 = 1) separated by distance 𝑟, the electric force is 𝐹⃗ = (𝑄1 𝑄2)/(4𝜋𝑟²𝜀0) (N)
• Charge 𝑄1 sets up a force field, producing a force on 𝑄2
• The electric field, 𝐸⃗⃗ , is a vector field defining the relationship between test charge 𝑄𝑡 and the force it experiences from source charge 𝑄1: 𝐸⃗⃗ = 𝐹⃗/𝑄𝑡
• For charge 𝑄1 producing a force on charge 𝑄2 (the test charge), the force is 𝐹⃗ = (𝑄1 𝑄2)/(4𝜋𝑟²𝜀0) (N)
• The electric field, 𝐸⃗⃗ , resulting from charge 𝑄1 is 𝐸⃗⃗ = 𝐹⃗/𝑄2= 𝑄1/(4𝜋𝑟²𝜀0) (V⁄m or N⁄C)
• Electric field lines extend radially from a point charge, with the direction being the force direction on a positive point charge
• Superposition of vector electric fields can find the resultant field at any given point

Potential Energy

• Assume 𝑄1 sets up electric field 𝐸⃗⃗ = 𝑄1/(4𝜋𝑟²𝜀0) (V⁄m)
• Positive test charge 𝑄𝑡 will be moved from an infinite radius to a point closer to positive charge 𝑄1
• Moving 𝑄𝑡 from infinity to 𝑟 = 𝑟𝐴 , work must be done against the repulsive force from field 𝐸⃗⃗ (Work Done = Force × Displacement)
• Force (𝐹⃗ = 𝐸⃗⃗ 𝑄𝑡 ) varies with radius because 𝐸⃗⃗ varies with radius, increasing in magnitude as 𝑟 decreases
• The force must be integrated from 𝑟 = ∞ to 𝑟 = 𝑟𝐴
• Work = Potential Energy is calculated as: Work = Potential Energy = − ∫(from ∞ to 𝑟𝐴) 𝐸⃗⃗ 𝑄𝑡 𝑑𝑟⃗
• Potential energy stored in the system, Potential Energy = − ∫(from ∞ to 𝑟𝐴) 𝐸⃗⃗ 𝑄𝑡 𝑑𝑟⃗ = − ∫(from ∞ to 𝑟𝐴) (𝑄1 𝑄𝑡)/(4𝜋𝑟²𝜀0) 𝑑𝑟⃗ = (𝑄1 𝑄𝑡)/(4𝜋𝑟𝐴 𝜀0) (J)

Electric Potential or Potential (V)

• Electric potential is the potential energy per unit test charge, Potential = (Potential Energy) / 𝑄𝑡
• Potential is calculated as:
  • 𝑉 = − ∫(from ∞ to 𝑟𝐴) 𝐸⃗⃗ 𝑄𝑡 𝑑𝑟⃗ / 𝑄𝑡 = − ∫(from ∞ to 𝑟𝐴) 𝐸⃗⃗ 𝑑𝑟⃗ = 𝑄1/(4𝜋𝑟𝐴 𝜀0) (V)
• Potential is measured in Volts (V), where Volts (V) = Joules (J) / Coulomb (C)
• Potential is a scalar quantity: use the principle of superposition

Potential Difference

• Potential difference is the difference in potential between two points
• Consider potential at points A and B
• Potential difference, 𝑉𝐵𝐴, is 𝑉𝐵𝐴 = 𝑉𝐵 − 𝑉𝐴 = − ∫(from ∞ to 𝑟𝐵) 𝐸⃗⃗ 𝑑𝑟⃗ − (− ∫(from ∞ to 𝑟𝐴) 𝐸⃗⃗ 𝑑𝑟⃗) = − [∫(from ∞ to 𝑟𝐵) 𝐸⃗⃗ 𝑑𝑟⃗ − ∫(from ∞ to 𝑟𝐴) 𝐸⃗⃗ 𝑑𝑟⃗] = − ∫(from 𝑟𝐴 to 𝑟𝐵) 𝐸⃗⃗ 𝑑𝑟⃗ (V)
• With 𝐸⃗⃗ = 𝑄1/(4𝜋𝑟²𝜀0), the potential difference is 𝑉𝐵𝐴 = − ∫(from 𝑟𝐴 to 𝑟𝐵) (𝑄1)/(4𝜋𝑟²𝜀0) 𝑑𝑟⃗ = (𝑄1/4𝜋𝜀0) [1/𝑟𝐵 − 1/𝑟𝐴] (V)
• An equipotential surface features constant electric potential: moving a charge requires no work
• Moving a charge in a closed path requires no work, with the start and finish potentials being the same (charge returns to its original point)

Electric Flux

• Electric flux, Φ𝐷 (Units: C), measures the electric field and represents a field's distribution in space
• 'Lines of electric flux' follow the electric field from positive to negative charges (or infinity if no other charges are present)
• Electric flux is primarily a means of visualising an electric field
• Electric fields are easily visualised and calculated for a single point charge as there are no complex interactions with other charges
• When two charges are considered, the situation becomes more complex, especially given that 1 coulomb of charge represents 6.24 × 10^18 electrons, so another approach is needed

Electric Flux Density

• Electric flux density, 𝐷⃗⃗ , describes the electric flux in a particular region of a system. More tightly packed ‘lines of flux’ represent a higher flux density
• Electric flux density, 𝐷⃗⃗ , is a vector quantity, like electric field strength, 𝐸⃗⃗
• Density and strength have the same direction but different magnitudes, with permittivity defining the relationship: 𝐷⃗⃗ = 𝜀0 𝜀𝑟 𝐸⃗⃗
• A higher flux density represents a stronger electric field, and flux density is independent of material properties, relating to charge and geometry
• The total electric flux through area A is given by the component of 𝐷⃗⃗ that is normal to the area of interest: Φ𝐷 = 𝐷⃗⃗ 𝐴 cos 𝜃 = 𝐷𝑛 𝐴

Gauss' Law

• Gauss' Law enables 𝐸⃗⃗ calculation for multiple charges/groups, stating, "the total electric flux out of a surface is equal to the net charge enclosed by that surface."
• Equation: ΦD = ∯ 𝐷⃗⃗ 𝑑𝑆⃗ = ∑ 𝑄𝑒𝑛𝑐
  • 𝑆⃗ is a closed surface
  • 𝐷⃗⃗ is the electric flux density
  • ΦD is the flux out of the surface
  • 𝑄 is the free charge enclosed by 𝑆⃗
• For symmetrical cases and an appropriate surface: ΦD = 𝐷𝑛 𝑆 = ∑ 𝑄𝑒𝑛𝑐, where 𝐷𝑛 is the flux density normal to the surface

Applying Gauss' Law

• To apply Gauss' Law, select a Gaussian surface, a closed surface generally selected to complement the system's geometry (though any closed surface works)
• Flux through each surface face is 'known' (easily defined)
• A spherical Gaussian surface suits a point charge, with all flux normal to the surface and evenly distributed
• A cylindrical Gaussian surface suits a long, straight conductor, with flux radial to the conductor and normal to the cylindrical surface

Gauss' Law Example

• Consider an isolated charge +𝑄 in free space (𝜀𝑟 = 1) and a spherical Gaussian surface of radius 𝑎
• Flux density is constant over the Gaussian surface; if not, consider small, defined sections and integrate
• Applying Gauss’ Law over the sphere (𝑆 = 4𝜋𝑎²) finds the flux density magnitude 𝐷, at radius 𝑎
  • ΦD = 4𝜋𝑎² 𝐷 = 𝑄 ⇒ 𝐷 = 𝑄/(4𝜋𝑎²)
• Electric field strength can then be found as 𝐸⃗⃗ = 𝐷/(𝜀0 𝜀𝑟) = 𝑄/(4𝜋𝜀0 𝑎²)
• The expression for electric field strength derived from Coulomb’s Law is the same as the one above from Gauss’ Law

Electric Fields Within Conductors

• Examining the relationship between electric field strength and flux density in free space and dielectrics, one should consider a conducting material
• Assume a conducting sphere of radius 𝑅 holds a charge +𝑄; what is the electric field strength within the sphere?
• Charges are free to move, so each exerts a force
• Balancing these forces requires the charge to spread evenly over the sphere's surface
• Select a spherical Gaussian surface (center matching the conducting sphere) with radius 𝑟 R (that is, the closed Gaussian surface is within the sphere)
• Apply Gauss' Law for the Gaussian surface: ΦD = ∯ 𝐷⃗⃗ 𝑑𝑆⃗ = ∑ 𝑄𝑒𝑛𝑐
  • 𝐷𝑛 𝑆 = ∑ 𝑄𝑒𝑛𝑐
  • 4𝜋𝑟² 𝐷 = ∑ 𝑄𝑒𝑛𝑐

Capacitance

• The charge enclosed by the Gaussian surface is zero since all charge is on the conducting sphere's surface
• Flux density and electric field strength are thus zero
• Capacitance is a system's ability to store energy in an electric field, defined as the charge stored per volt (potential difference)
  • Capacitance (𝐶) = 𝑄/𝑉, measured in farads (F) or Coulombs per Volt (C/V)
• Applying Gauss’ Law for a spherical Gaussian surface around a point charge shows that at radius 𝑎, 𝐸 = 𝑄/(4𝜋𝜀0 𝑎²)
• Assume the Gaussian surface is a physical conducting sphere of radius 𝑎
• The potential of its surface can be calculate as: 𝑉 = − ∫(from ∞ to 𝑎) 𝐸⃗⃗ 𝑑𝑟⃗ = 𝑄/(4𝜋𝜀0 𝑎)
  • Note: same as potential difference between the sphere and a point at an infinite radius
• Capacitance depends only on the physical properties (dimensions and materials) of the system, not the charge
  • 𝐶 = 𝑄/𝑉 = 4𝜋𝜀0 𝑎
• The capacitance of any system depends only on its physical properties and not any charge it might hold

Parallel Plate Capacitor

• Assume two parallel conducting plates separated by air (free space), at distance 𝑑 with cross-sectional area 𝐴
• Each plate holds charge of magnitude 𝑄, but of opposite polarity
• Neglecting fringing at the edges, find the system's capacitance
• A Gaussian surface can be drawn along lines of "known” flux density
• If the total charge on each plate is 𝑄, the charge density, 𝑞, is 𝑞 = 𝑄/𝐴 (C⁄m²)
• Flux normal to the Gaussian surface is zero for all faces except between the plates (area 𝑑𝐴)
• By Gauss’ Law, the flux, ΦD , out of the surface equals the enclosed charge: ΦD = 𝐷 𝑑𝐴 = 𝑞 𝑑𝐴
• Flux density (𝐷) between the plates equals charge density (𝑞): 𝐷 = 𝑞 = 𝑄/𝐴
• Electric field strength (𝐸) by definition is 𝐸 = 𝐷/𝜀0 = 𝑄/(𝐴𝜀0) = 𝑞/𝜀0
• As with the spherical example, the electric field depends on the system's dimensions and the charge it holds
• Potential difference between the plates is found as 𝑉 = − ∫(from 𝑑 to 0) 𝐸 𝑑𝑥 = 𝑄𝑑/(𝐴𝜀0) = 𝑞𝑑/𝜀0
• Capacitance is the charge per volt (𝐶 = 𝑄/𝑉) is found through 𝐶 = 𝑄/𝑉 = 𝐴𝜀0/𝑑
• As with the spherical example, capacitance depends solely on the physical properties of the system (its dimensions and materials)

Energy Stored in a Capacitor

• The work done (and therefore energy stored) charging a capacitor can be found by integrating the voltage across the capacitor with respect to the charge held on the plates
• For known capacitance, voltage varies linearly with charge: 𝑉(𝑄) = 𝑄/𝐶
• If the maximum charge is 𝑄𝑐, the energy stored is: Energy stored = ∫(from 0 to 𝑄𝑐) 𝑉(𝑄)𝑑𝑄 = ∫(from 0 to 𝑄𝑐) 𝑄/𝐶 𝑑𝑄 = 1/2 (𝑄𝑐²/𝐶)
• Rewriting depending on which variables are known:
  • Energy stored = 1/2 (𝑄²/𝐶) = 1/2 (𝑄𝑉) = 1/2 (𝐶𝑉²) (J)

Dielectric Materials

• As shown, capacitance depends solely on a system's physical properties (dimensions, materials)
• A parallel capacitor in free space is constrained by plate area (𝐴) and separation distance (𝑑)
  • 𝐶 = 𝐴𝜀0/𝑑
• Increasing plate area or reducing distance increases capacitance
• Alternatively, change the material between plates to change electric field strength (flux density remains constant as it depends on charge)
• Relative permittivity (𝜀𝑟) describes a material's permittivity relative to free space: 𝐸⃗⃗ = 𝐷⃗⃗ /(𝜀0 𝜀𝑟)
• Increasing capacitance increases the energy stored in the system: Energy stored = 1/2 (𝐶𝑉²)
• Applying an electric field to a classical atomic model polarises the atom
• A dielectric material is an insulator (no charge flows). With zero electric field, polarised molecules balance, giving a net electric field of zero. When a field is applied, dipoles align, producing polarised material

Applying Gauss' Law with Dielectrics

• Assuming the same parallel plate capacitor, with free space replaced by dielectric material of relative permittivity 𝜀𝑟
• Gauss’ Law application is unaffected by a dielectric; a Gaussian surface is selected such that flux passes through area 𝑑𝐴
• 𝐷 = 𝐷 𝑑𝐴 = 𝑞 𝑑𝐴, where 𝑞 is the charge density (C/m²)
• Flux density depends on charge on the plates and is unaffected by the dielectric: 𝐷 = 𝑄/𝐴 = 𝑞
• The dielectric opposes the field, and reduces it compared to the case without the dielectric: 𝐸 = 𝐷/(𝜀0 𝜀𝑟) = 𝑄/(𝐴𝜀0 𝜀𝑟) = 𝑞/(𝜀0 𝜀𝑟)
• Potential difference is found as 𝑉 = − ∫(from 𝑑 to 0) 𝐸 𝑑𝑥 = 𝑄𝑑/(𝐴𝜀0 𝜀𝑟) = 𝑞𝑑/(𝜀0 𝜀𝑟)
• Capacitance is 𝐶 = 𝑄/𝑉 = (𝐴𝜀0 𝜀𝑟)/𝑑
• The presence of a dielectric has reduced the electric field strength but increased the capacitance and the energy that can be stored
• Common dielectrics include:
  • Air: 1.00054
  • Paper: 3.5
  • Mica: 3 – 6
  • Aluminium oxide: 8.5
  • Tantalum dioxide: 86 – 173
  • Ceramics: Up to order of 1000s

Electric Stress and Breakdown Voltage

• Electric field strength and potential for a parallel plate capacitor example are:
  • 𝐸 = 𝑄/(𝐴𝜀0 𝜀𝑟)
  • 𝑉 = 𝑄𝑑/(𝐴𝜀0 𝜀𝑟)
• By substitution, the relationship between electric field strength (𝐸) and applied voltage (𝑉) is 𝑉 = 𝐸𝑑, where 𝑑 is the separation between the charged plates
• Every dielectric has a dielectric strength (𝐸𝑑𝑠), the maximum allowable electric field that it withstands without breaking down and becoming a conductor
• Typical dielectric strengths (𝐸𝑑𝑠) include:
  • Air: 3 × 10^6 V/m
  • Paper: 16 × 10^6 V/m
  • Waxed paper: 36 × 10^6 V/m
  • Distilled water: 70 × 10^6 V/m
  • Mica: 100 × 10^6 V/m
  • Diamond: 2000 × 10^6 V/m
• The breakdown voltage (𝑉𝑏) of a dielectric is calculated based on strength and thickness of the layer
• A parallel plate capacitor with 0.5 mm separation and air has a breakdown voltage of 1.5 kV
  • Putting paper (strength 16 × 10^6 V/m)between the plates allows the max voltage to increase to 8 kV because paper has higher dielectric strength
  • Calculating breakdown voltages:
    • 𝑉𝑏 𝑎𝑖𝑟 = 𝐸𝑑𝑠𝑎𝑖𝑟 𝑑 = 3 × 10^6 × 0.5 × 10^(−3) = 1500 V
    • 𝑉𝑏 𝑝𝑎𝑝𝑒𝑟 = 𝐸𝑑𝑠𝑝𝑎𝑝𝑒𝑟 𝑑 = 16 × 10^6 × 0.5 × 10^(−3) = 8000 V
• The exact relationship between 𝑉 and 𝐸 matches the system geometry
• The maximum electric field strength in the system should not exceed the materials' dielectric strengths
• For example, electrical cables have a max permitted voltage depending on the chosen insulating materials