Question

Space between plates of a parallel plate capacitor is filled with three
dielectric slab as shown in figure. Find equivalent dielectric constant for the
arrangement shown. Area of each plate is A.
​

Answer 1

Answer:

k=1      k=2

Explanation:

Answer 2

The equivalent dielectric constant for the arrangement shown is 6.67.

Given,

3 dielectric slabs between parallel plate capacitors

k₁ = 3, k₂ = 12, k₃ = 6

d₁ = d₂ = d₃ = d/2

Area of each plate = A

A₁ = A₂ = A/2

A₃ = A

To Find,

Equivalent dielectric constant (k)

Solution,

Let the equivalent dielectric constant be k

Thus, Capacitance for the system is -

C = kε₀(A/d)

Now, k₁ and k₂ are in parallel combination → C₁ and C₂ in parallel

C₁ = k₁ε₀(A₁/d₁)

C₁ = 3*ε₀({A/2}/{d/2})

C₁ = 3ε₀(A/d)

C₂ = k₂ε₀(A₂/d₂)

C₂ = 12*ε₀({A/2}/{d/2})

C₂ = 12ε₀(A/d)

For C₁ and C₂ in parallel, the C equivalent is given as -

C' = C₁ + C₂

C' = 3ε₀(A/d) + 12ε₀(A/d)

C' = 15ε₀(A/d)

Now C' and C₃ are in series -

C₃ = k₃ε₀(A₃/d₃)

C₃ = 6*ε₀(A/{d/2})

C₃ = 12ε₀(A/d)

Now C equivalent for series is given as -

$\frac{1}{C} = \frac{1}{C'} + \frac{1}{C3}$

C = (C' * C₃) / (C' + C₃)

kε₀(A/d) = {15ε₀(A/d) * 12ε₀(A/d)} / {15ε₀(A/d) + 12ε₀(A/d)}

Cancelling like terms we get,

k = (15 * 12) / (15 + 12)

k = 180/27

k = 6.67

Therefore, the equivalent dielectric constant is 6.67.

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