(a) Surface Charge Density of Polarization Charges ($\sigma_p$)

Consider the region between the capacitor plates. We have a composite dielectric system: a belt of thickness $h$ with dielectric constant $\varepsilon_r$, and an air gap of thickness $d-h$.

Let the electric field inside the belt be $E_{in}$ and the field in the air gap be $E_{air}$. From the continuity of electric displacement field $D$ (since there are no free charges on the belt surface initially):

$$ D = \varepsilon_0 E_{air} = \varepsilon_0 \varepsilon_r E_{in} \implies E_{air} = \varepsilon_r E_{in} $$

The total potential difference $V$ across the plates is the sum of the potential drops:

$$ V = E_{air}(d-h) + E_{in}h $$

Substituting $E_{air} = \varepsilon_r E_{in}$:

$$ V = \varepsilon_r E_{in}(d-h) + E_{in}h = E_{in} [\varepsilon_r(d-h) + h] $$ $$ E_{in} = \frac{V}{\varepsilon_r(d-h) + h} $$

The polarization $P$ is given by $P = \varepsilon_0 (\varepsilon_r – 1) E_{in}$. The surface charge density of bound (polarization) charges $\sigma_p$ is equal to the magnitude of $P$.

(b) Charge $q$ on the Dome after time $t$

The belt has a width $b$ and moves with velocity $v$. The area of the belt entering the dome per unit time is $dA/dt = bv$.

The brush $B_1$ collects the polarization charges from the outer surface. The rate at which charge is delivered to the dome (current $I_{supply}$) is:

$$ I_{supply} = \sigma_p \times \frac{dA}{dt} = \sigma_p b v $$

Assuming no leakage initially, the total charge $q$ after time $t$ is:

(c) Current $I$ shown by the Ammeter

The ammeter is connected to brush $B_2$, which collects charges from the inner surface of the belt as it leaves the dome. Due to the polarization of the dielectric, if the outer surface has charge density $+\sigma_p$, the inner surface has $-\sigma_p$.

The magnitude of the current flowing to the ground is the rate at which this charge is removed:

(d) Steady State Charge with Ohmic Leakage

In the steady state, the rate of charge arriving at the dome equals the rate of charge leaking into the air. Let the air have resistivity $\rho$.

Rate of charge entering: $I_{in} = \sigma_p b v$.

Rate of charge leaking ($I_{leak}$): Using the relation between current density $J$ and Electric Field $E$ for an ohmic medium, $J = E/\rho$. The leakage current is the flux of $J$ over a closed surface enclosing the dome:

$$ I_{leak} = \oint \vec{J} \cdot d\vec{A} = \oint \frac{\vec{E}}{\rho} \cdot d\vec{A} = \frac{1}{\rho} \oint \vec{E} \cdot d\vec{A} $$

From Gauss’s Law, $\oint \vec{E} \cdot d\vec{A} = q_{enclosed} / \varepsilon_0 = q / \varepsilon_0$.

$$ I_{leak} = \frac{q}{\rho \varepsilon_0} $$

Equating $I_{in} = I_{leak}$ for steady state:

$$ \sigma_p b v = \frac{q}{\rho \varepsilon_0} \implies q = \rho \varepsilon_0 \sigma_p b v $$

Substituting $\sigma_p$ from part (a):

(e) Steady State Charge with Corona Discharge

Here, the leakage current density is non-linear: $j = \frac{E}{\rho} + \beta E^2$.

The total leakage current $I_{out}$ is the integral of $j$ over the surface area of the spherical dome ($4\pi R^2$). Assuming the field $E$ is uniform over the sphere’s surface, $E = \frac{q}{4\pi \varepsilon_0 R^2}$.

$$ I_{out} = \oint \left( \frac{E}{\rho} + \beta E^2 \right) dA = \frac{1}{\rho}\oint E dA + \beta \oint E^2 dA $$ $$ I_{out} = \frac{q}{\rho \varepsilon_0} + \beta \left( \frac{q}{4\pi \varepsilon_0 R^2} \right)^2 (4\pi R^2) $$ $$ I_{out} = \frac{q}{\rho \varepsilon_0} + \frac{\beta q^2}{4\pi \varepsilon_0^2 R^2} $$

At steady state, $I_{in} = I_{out}$. Let $I_{in} = \sigma_p b v$.

$$ \frac{\beta}{4\pi \varepsilon_0^2 R^2} q^2 + \frac{1}{\rho \varepsilon_0} q – I_{in} = 0 $$

This is a quadratic equation in $q$ of the form $Ax^2 + Bx – C = 0$, where $A = \frac{\beta}{4\pi \varepsilon_0^2 R^2}$, $B = \frac{1}{\rho \varepsilon_0}$, and $C = I_{in}$. The solution is:

$$ q = \frac{-B + \sqrt{B^2 + 4AC}}{2A} = \frac{B}{2A} \left( \sqrt{1 + \frac{4AC}{B^2}} – 1 \right) $$

Calculating the pre-factor $\frac{B}{2A}$:

$$ \frac{B}{2A} = \frac{1/(\rho \varepsilon_0)}{2\beta / (4\pi \varepsilon_0^2 R^2)} = \frac{1}{\rho \varepsilon_0} \cdot \frac{2\pi \varepsilon_0^2 R^2}{\beta} = \frac{2\pi \varepsilon_0 R^2}{\rho \beta} $$

Calculating the term inside the square root $\frac{4AC}{B^2}$:

$$ \frac{4AC}{B^2} = \frac{4 \cdot (\beta / 4\pi \varepsilon_0^2 R^2) \cdot I_{in}}{1 / (\rho^2 \varepsilon_0^2)} = \frac{\beta I_{in}}{\pi \varepsilon_0^2 R^2} \cdot \rho^2 \varepsilon_0^2 = \frac{\rho^2 \varepsilon_0 \beta}{\pi R^2} \left( \frac{I_{in}}{\varepsilon_0} \right) $$

Substituting $I_{in} = \sigma_p b v$ and the expression for $\sigma_p$: