Step-by-Step Derivation

1. Electric Field Distribution

Let the potential difference applied between the plates be $V$. The space between the plates consists of:

• A dielectric slab of thickness $t$ with permittivity $\varepsilon = \varepsilon_0 \varepsilon_r$.
• Air gaps (or vacuum) occupying the remaining distance $d – t$.

Let the surface charge density on the plates be $\sigma_{free}$. The electric displacement $D$ is uniform throughout the region (neglecting fringe effects) and is equal to $\sigma_{free}$.

$$ D = \sigma_{free} $$

The electric field in the air gap ($E_0$) and in the dielectric ($E$) are related by:

$$ E_0 = \frac{D}{\varepsilon_0} = \frac{\sigma_{free}}{\varepsilon_0} $$ $$ E = \frac{D}{\varepsilon_0 \varepsilon_r} = \frac{\sigma_{free}}{\varepsilon_0 \varepsilon_r} = \frac{E_0}{\varepsilon_r} $$

The potential difference $V$ is the sum of potential drops across the air and the dielectric:

$$ V = E_0 (d – t) + E t $$ Substitute $E_0 = \varepsilon_r E$: $$ V = (\varepsilon_r E)(d – t) + E t $$ $$ V = E [\varepsilon_r(d – t) + t] $$

From this, we can express the electric field inside the dielectric, $E$, in terms of the applied voltage $V$:

$$ E = \frac{V}{\varepsilon_r(d – t) + t} $$ Or, we can define effective distance $d’ = \varepsilon_r(d-t) + t$, so $E = V/d’$.

2. Mechanical Stress on the Dielectric

We need to identify the force per unit area (stress) tending to rupture the slab. The rupture is caused by the mechanical stress induced by the electric field. For a dielectric slab between plates, there is an electrostatic force acting on the polarized charges.

Consider the energy density or the force calculation. The electrostatic stress pulling the surfaces of the dielectric outwards (tensile stress) is given by the formula:

$$ \text{Stress } S = \frac{1}{2} \varepsilon_0 (\varepsilon_r – 1) E_{inside}^2 $$

Alternatively, consider the force density $f = (P \cdot \nabla)E$. Since the field is uniform, bulk force is zero. The forces act on the interfaces. The stress experienced by the dielectric due to the field is:

$$ \text{Stress} = \frac{1}{2} \varepsilon E^2 – \frac{1}{2} \varepsilon_0 E^2 = \frac{1}{2} \varepsilon_0 \varepsilon_r E^2 – \frac{1}{2} \varepsilon_0 E^2 $$ Wait, this formula $f = \frac{1}{2}(\epsilon – \epsilon_0) E^2$ usually refers to the force density on a liquid dielectric surface. Let’s look at the forces on the faces of the slab.

The induced surface charge density on the dielectric face is $\sigma_{ind} = – P = – \varepsilon_0 (\varepsilon_r – 1) E$. The field acting on this induced charge is the average of the field inside and outside. $E_{avg} = \frac{E_{inside} + E_{outside}}{2} = \frac{E + \varepsilon_r E}{2} = \frac{E(1+\varepsilon_r)}{2}$. Force per unit area $F/A = \sigma_{ind} E_{avg}$. This seems complicated. Let’s use the standard expression for the Maxwell stress tensor or the energy method.

Standard Result: The mechanical stress (tensile) on a dielectric boundary normal to the field is:

$$ \sigma_{stress} = \frac{1}{2} \varepsilon_0 (\varepsilon_r – 1) \varepsilon_r E_{inside}^2 $$ Wait, let’s check the boundary conditions. $D_n$ is continuous. $D_{air} = D_{diel} = \varepsilon_0 \varepsilon_r E$. Field in air $E_{air} = \varepsilon_r E$. Stress pulling the surface = Energy density diff? $F/A = \frac{1}{2} \varepsilon_0 E_{air}^2 – \frac{1}{2} \varepsilon E^2 = \frac{1}{2} \varepsilon_0 (\varepsilon_r E)^2 – \frac{1}{2} (\varepsilon_0 \varepsilon_r) E^2$ $F/A = \frac{1}{2} \varepsilon_0 \varepsilon_r^2 E^2 – \frac{1}{2} \varepsilon_0 \varepsilon_r E^2$ $F/A = \frac{1}{2} \varepsilon_0 \varepsilon_r (\varepsilon_r – 1) E^2$. This is the net outward force per unit area on the dielectric surface. This stress tends to stretch or rupture the slab. So, the breaking stress $\sigma_b$ must be equated to this electrostatic stress. $$ \sigma_b = \frac{1}{2} \varepsilon_0 \varepsilon_r (\varepsilon_r – 1) E^2 $$

3. Calculating the Voltage

We have the condition for rupture:

$$ \sigma_b = \frac{1}{2} \varepsilon_0 \varepsilon_r (\varepsilon_r – 1) E^2 $$

Solving for the electric field $E$ inside the dielectric:

$$ E^2 = \frac{2 \sigma_b}{\varepsilon_0 \varepsilon_r (\varepsilon_r – 1)} $$ $$ E = \sqrt{\frac{2 \sigma_b}{\varepsilon_0 \varepsilon_r (\varepsilon_r – 1)}} $$

Now, substitute this back into the potential equation derived in Step 1:

$$ V = E [\varepsilon_r(d – t) + t] $$ $$ V = \left( \sqrt{\frac{2 \sigma_b}{\varepsilon_0 \varepsilon_r (\varepsilon_r – 1)}} \right) [\varepsilon_r(d – t) + t] $$

Rearranging the terms for clarity:

$$ V = [\varepsilon_r d – \varepsilon_r t + t] \sqrt{\frac{2 \sigma_b}{\varepsilon_0 \varepsilon_r (\varepsilon_r – 1)}} $$ $$ V = [\varepsilon_r d – t(\varepsilon_r – 1)] \sqrt{\frac{2 \sigma_b}{\varepsilon_0 \varepsilon_r (\varepsilon_r – 1)}} $$