Irodov 3.160: Leakage Read Off the Capacitance – Amit & Deepali Sinha Foundation

Problem Irodov 3.160

A parallel-plate capacitor of capacitance $C = 4.0$ nF has the space between its plates filled with glass of resistivity $\rho = 100$ G $\Omega\cdot$ m and permittivity $\varepsilon = 6.0$ . A voltage $V = 2.0$ kV is applied across it. What current leaks through the glass?

The parallel-plate capacitor edge on, its gap d filled with glass of resistivity ρ and permittivity ε, with a lead from each plate for the applied voltage. — The parallel-plate capacitor edge on, its gap $d$ filled with glass of resistivity $\rho$ and permittivity $\varepsilon$ , with a lead from each plate for the applied voltage.

Given

$\rho = 100$ G $\Omega\cdot$ m
$C = 4.0$ nF
$V = 2.0$ kV
glass, $\varepsilon = 6.0$ (table value)
edge effects neglected (assumed)

Find

the leakage current

I

Solution

Step 1The resistance of the glass layer

Call the plate area $S$ and the gap $d$ ; neither is given, and both will cancel. Neglecting the edges, the current crosses the gap uniformly, so the glass is a conductor of length $d$ and cross-section $S$ :

R = \frac{\rho\, d}{S} \tag{1}

Step 2The capacitance holds the same geometry

With the glass, of permittivity $\varepsilon = 6.0$ (a table value), filling the gap, the capacitance is

C = \frac{\varepsilon_0\varepsilon S}{d} \tag{2}

Multiplying (1) by (2), the ratio $d/S$ cancels:

RC = \frac{\rho\, d}{S}\cdot\frac{\varepsilon_0\varepsilon S}{d} = \varepsilon_0\varepsilon\rho

Step 3The leakage current

The leakage current is $I = V/R$ , and dividing the product $RC$ by $C$ gives the resistance:

R = \frac{\varepsilon_0\varepsilon\rho}{C}

Numerically, the material time and then the current are

\begin{aligned} \varepsilon_0\varepsilon\rho &= 8.85\times10^{-12}\cdot 6.0\cdot 1.0\times10^{11}\ \text{s} = 5.31\ \text{s}\\ I &= \frac{2.0\times10^{3}\cdot 4.0\times10^{-9}}{5.31}\ \text{A} \end{aligned}

I = \frac{V C}{\varepsilon_0\varepsilon\rho} \approx 1.5\ \mu\text{A}

(3)

Check

Dimensions: $\varepsilon_0\varepsilon\rho$ is (F/m)( $\Omega\cdot$ m), that is F $\cdot\Omega$ = s, and $VC$ is a charge in coulombs, so their ratio is in amperes. Limits: a perfect insulator, $\rho \to \infty$ , gives $I \to 0$ , an ideal capacitor; and at fixed $C$ the leakage falls as $1/\varepsilon$ , because with a higher permittivity the same capacitance needs a smaller ratio $S/d$ , which leaves $R = \rho d/S$ larger.

Remark

The time $\varepsilon_0\varepsilon\rho \approx 5$ s is the self-discharge time of this capacitor: disconnected, it loses its charge as $e^{-t/RC}$ whatever its size or shape. How long a capacitor holds its charge is a property of its dielectric alone.