Question 6.12: Determine the capacitance of each of the capacitors in Figur...

(a) Since D and E are normal to the dielectric interface, the capacitor in Figure 6.20(a) can be treated as consisting of two capacitors C_{1} and C_{2} in series as in Figure 6.16(a).

C_{1}=\frac{\varepsilon _{o}\varepsilon _{r1}S}{d/2}=\frac{2\varepsilon_{o}\varepsilon_{r1}S}{d},       C_{2}=\frac{2\varepsilon_{o}\varepsilon_{r2}S}{d}

The total capacitor C is given by

C=\frac{C_{1}C_{2}}{C_{1}+C_{2}}=\frac{2\varepsilon_{o}S}{d}\frac{(\varepsilon_{r1}\varepsilon_{r2})}{\varepsilon_{r1}+\varepsilon_{r2}}

=2\cdot\frac{10^{-9}}{36\pi}\cdot\frac{30\times10^{-4}}{5\times10^{-3}}\cdot\frac{4\times6}{10}                                                                                  (6.12.1)

C=25.46 pF

(b) In this case, D and E are parallel to the dielectric interface. We may treat the capacitor as consisting of two capacitors C_{1} and C_{2} in parallel (the same voltage across C_{1} and C_{2}) as in Figure 6.16(b).

C_{1}=\frac{\varepsilon _{o}\varepsilon _{r1}S/2}{d}=\frac{\varepsilon_{o}\varepsilon_{r1}S}{2d},       C_{2}=\frac{\varepsilon_{o}\varepsilon_{r2}S}{2d}

The total capacitance is

C=C_{1}+C_{2}=\frac{\varepsilon_{o}S}{2d}(\varepsilon_{r1}+\varepsilon_{r2})

=\frac{10^{-9}}{36\pi}\cdot\frac{30\times10^{-4}}{2\cdot(5\times10^{-3})}\cdot10                                                                               (6.12.2)

C=26.53 pF

Notice that when \varepsilon_{r1}=\varepsilon_{r2}=\varepsilon_{r}, eqs. (6.12.1) and (6.12.2) agree with eq. (6.22)

C=\frac{Q}{V}=\frac{\varepsilon S}{d}

as expected.