D2-3. Properties of Charge and Permittivity

1. Charge Properties, Conductors & Permittivity

Quantisation & Conservation of Charge

  • Quantisation: Charge always exists as an integer multiple of the elementary charge ($e = 1.60 \times 10^{-19}\text{ C}$). A particle cannot have a fraction of $e$.
  • Conservation: In an isolated system, the total net charge remains constant during any process.

Relative Permittivity (Dielectric Constant)

For a given material, relative permittivity is defined as the ratio of the permittivity of a material to the permittivity of free space. It can be expressed as:

$$\varepsilon_r = \frac{\varepsilon}{\varepsilon_0}$$

where F means Farad and:

  • $\varepsilon_r$ = relative permittivity
  • $\varepsilon$ = permittivity of a material ($\text{F m}^{-1}$)
  • $\varepsilon_0$ = permittivity of free space ($\text{F m}^{-1}$)

If a material is placed between charges, the Coulomb constant becomes $$ k = \frac{1}{4\pi\varepsilon}$. Because $\varepsilon > \varepsilon_0, $$ this makes the resulting electric field weaker than it would be in a vacuum.

Vacuum ($\varepsilon = \varepsilon_0$)

+ -

Dense Lines = Strong Field

Effective k is Large

Dielectric Material ($\varepsilon > \varepsilon_0$)

Material (ε) + -

Fewer Lines = Weaker Field

Effective k is Smaller

Electric Field Inside a Conductor

  • When a solid or hollow conducting sphere is charged, all excess charge distributes evenly on the outer surface due to mutual repulsion.
  • Crucial Fact: The electric field inside the conductor is exactly zero ($\vec{E} = 0$). Outside, it behaves exactly like a point charge.
+ + + + + + + + E = 0 (Inside) Field lines only exist outside the surface

2. Examples

Example 1

Problem: Two point charges of equal charge are separated in free space by a distance $d$ and experience an electric force $F_{free\ space}$. The same point charges are now separated by a wall of concrete $d/2$ thick and experience an electric force $F_{concrete}$.

d 2

The permittivity of concrete is four times that of free space. What is $$ \frac{F_{concrete}}{F_{free\ space}} \ ? $$

Solution:

The force between the charges in free space is $$ F_{free\ space} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{d^2} $$ When the charges are separated by a concrete wall of thickness $d/2$, their new separation distance is exactly $d/2$, and the space between them is completely filled by a medium with a permittivity of $\varepsilon = 4\varepsilon_0$. The new force is calculated as $$ F_{concrete} = \frac{1}{4\pi(4\varepsilon_0)} \frac{q_1 q_2}{(d/2)^2} $$ Squaring the halved distance yields a factor of $1/4$ in the denominator, which perfectly cancels out the factor of $4$ from the concrete's permittivity. As a result, the two effects multiply to 1, leaving $F_{concrete} = F_{free\ space}$.