Resistance vs. Capacitance

a) Consider two conductors of some shape. Use them in two alternative ways, as a capacitor and as a resistor as shown in (a) and (b) of Figure 1.1, respectively. In case (a), the space between the conductors is filled with a homogeneous material of permittivity ε, while in case (b), it is filled with a homogeneous material of finite conductivity σ. By considering separately these two cases, prove the relation

Figure 1.1

between the capacitance C in case (a) and the resistance R in case

(b). If you cannot give a general proof, try at least some special example, e.g., parallel plates.

b) Two conducting spheres have their centers a distance apart. Their respective radii are a and b. Show that when c >> a, b the capacitance of this system will be given approximately by

c) Two small, spherical, perfectly conducting electrodes of radii a and b are embedded in an infinite medium of conductivity σ. Their centers are separated by a distance c >> a, b Find the resistance between them without using (a) and (b).

Hint: If two electrodes at potentials V₁ and V₂ are embedded in a medium of finite conductivity, the currents I₁ and I₂ leaving each of them are related to the potentials by the formulae V₁ = R₁₁ I₁ + R₁₂ I₂, V₂ = R₂₁ I₁ + R₂₂ I_2. Determine the coefficients R_ij by considering cases with I₂ = 0 and I₁ = 0.

d) Check the results of (b) and (c) by using (a).

SOLUTION

a) Enclose one of the conductors in a surface and use Maxwell’s equation (see Figures 1.2a and 1.2b)

So

(1)

Take the volume integral of (1) and transform into a surface integral:

(2)

Figures 1.2a, b

Using this result and the definition of C, we find

which yields

(3)

Now treat the resistor problem by starting with J = σE. Finding the current flux through the same surface,

So

(4)

Equating (3) and (4), we find

and finally

(5)

The parallel plate capacitor has the following capacitance and resistance:

Thus, we confirm the general result for RC.

b) Find the potential at the surface of each conductor:

(6)

So

(7)

c) Following the hint, consider the potential drop from each conductor to infinity when first I₂ = 0, and then I₁ = 0:

so

In the same fashion

Now, V₁ = (R₁₁ – R₁₂)I and V₂ = (R₂₁ – R₂₂)I, where I = I₁ = -I₂. So we obtain for R

(8)

d) Multiplying (7) by (8), we find the result in (5) with ε = 1

مواضيع ذات صلة

مصطلحات الفيزياء النووية

إزاحة حمراء REDSHIFT

المركبات النانوية (Nanocomposites)

مولد فان دي جراف الالكتروستاتي

التطويرات التقنية الحديثة في المعجلات الالكتروستاتية

معجلات الالكترونات الالكتروستاتية

معجلات الفان دي جراف الترادفية

الحد من زيادة الطاقة في معجل فان دي جراف

أنبوبة التعجيل قي معجلات الجسيمات من نوع فان دي جراف

ارتكاز معجل فان دي جراف

مبدأ تشغيل معجل الفان دي جراف

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