Question 25.48: Figure 25-47 shows a parallel-plate capacitor with a plate a......

Fundamentals of Physics Extended – Instructor’s Solutions Manual [1360823]

Figure 25-47 shows a parallel-plate capacitor with a plate area A = 5.56 cm² and separation d = 5.56 mm. The left half of the gap is filled with material of dielectric constant $\kappa_1$ = 7.00; the right half is filled with material of dielectric constant $\kappa_2$ = 12.0.What is the capacitance?

1360823-Figure 25.47

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The capacitor can be viewed as two capacitors C $_1$ and C $_2$ in parallel, each with surface area A/2 and plate separation d, filled with dielectric materials with dielectric constants $\kappa_1$ and $\kappa_2$ , respectively. Thus, (in SI units),

$\begin{aligned} C & =C_1+C_2=\frac{\varepsilon_0(A / 2) \kappa_1}{d}+\frac{\varepsilon_0(A / 2) \kappa_2}{d}=\frac{\varepsilon_0 A}{d}\left(\frac{\kappa_1+\kappa_2}{2}\right) \\ & =\frac{\left(8.85 \times 10^{-12} \,C ^2 / N \cdot m ^2\right)\left(5.56 \times 10^{-4} \,m ^2\right)}{5.56 \times 10^{-3} \,m }\left(\frac{7.00+12.00}{2}\right)=8.41 \times 10^{-12} \,F . \end{aligned}$

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