The space between the plates of a parallel plate capacitor of capacitance C (without any dielectric) is now filled with three dielectric slabs of dielectric constants $$K_{1}=2,K_{2}=3\text{ and } K_{3}=5$$ (as shown in figtue). lf new capacitance is $$\frac{n}{3}C$$ then the value of n is_____.

Let original capacitance (no dielectric):

$$C=\frac{\varepsilon_0A}{d}$$

step 1: understand geometry

• Top half (thickness d/2): dielectric $$K_1=2$$, area A
• Bottom half (thickness d/2): split into two equal areas:

• left: $$K_2=3$$, area A/2
• right: $$K_3=5$$, area A/2

step 2: bottom part (parallel combination)

Each has thickness d/2:

$$C_2=\frac{3\varepsilon_0(A/2)}{d/2}=\frac{3\varepsilon_0A}{d}$$

$$C_3=\frac{5\varepsilon_0(A/2)}{d/2}=\frac{5\varepsilon_0A}{d}$$

Parallel:

$$C_{bottom}=C_2+C_3=\frac{8\varepsilon_0A}{d}$$

step 3: top part

$$C_{top}=\frac{2\varepsilon_0A}{d/2}=\frac{4\varepsilon_0A}{d}$$

step 4: series combination

Top and bottom are in series:

$$\frac{1}{C'}=\frac{1}{C_{top}}+\frac{1}{C_{bottom}}=\frac{d}{4\varepsilon_0A}+\frac{d}{8\varepsilon_0A}$$

$$=\frac{3d}{8\varepsilon_0A}$$

$$C'=\frac{8\varepsilon_0A}{3d}$$