A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness $d_{1}$ and dielectric constant $k_{1}$ and the other has thickness $d_{2}$ and dielectric constant $k_{2}$ as shown in Fig. 2.3. This arrangement can be thought as a dielectric slab of thickness $d (= d_{1} + d_{2})$ and effective dielectric constant $k$ . The $k$ is

\frac{k_{1} d_{1} + k_{2} d_{2}}{k_{1} + k_{2}}

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\frac{2 k_{1} k_{2}}{k_{1} + k_{2}}

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\frac{k_{1} d_{1} + k_{2} d_{2}}{d_{1} + d_{2}}

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\frac{k_{1} k_{2} (d_{1} + d_{2})}{(k_{1} d_{1} + k_{2} d_{2})}

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Solution

The correct option is D $\frac{k_{1} k_{2} (d_{1} + d_{2})}{(k_{1} d_{1} + k_{2} d_{2})}$
Step 1: Draw the diagram of the given problem.

Step 2: Calculate the equivalent capacitance.

From the figure system can be considered as two capacitors $C_{1}$ and $C_{2}$ connected in series as shown in figure.

$C_{1} = \frac{k_{1} ε_{0} A}{d_{1}}, C_{2} = \frac{k_{2} ε_{0} A}{d_{2}}$

$\frac{1}{C_{e q}} = \frac{1}{C_{1}} + \frac{1}{C_{2}}$

$\Rightarrow C_{e q} = \frac{C_{1} C_{2}}{C_{1} + C_{2}}$

$C_{e q} = \frac{\frac{k_{1} ε_{0} A}{d_{1}} \frac{k_{2} ε_{0} A}{d_{2}}}{\frac{k_{1} ε_{0} A}{d_{1}} + \frac{k_{2} ε_{0} A}{d_{2}}} = \frac{k_{1} k_{2} ε_{0} A}{k_{1} d_{2} + k_{2} d_{1}}$ ..(i)

We can write the equivalent capacitance as

$C = \frac{K ε_{0} A}{d_{1} + d_{2}}$ …(ii)

Step 3: Fine the effective dielectric constant.

On comparing (i) and (ii) we have

$k = \frac{k_{1} k_{2} (d_{1} + d_{2})}{k_{1} d_{2} + k_{2} d_{1}}$

Final Answer: (c)

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Similar questions

Q. A parallel plate capacitor with air between the plates has a capacitance of 9

p F

. The separation between its plates is

d

. The space between the plates is now filled with two dielectrics. One of the dielectric has dielectric constant

K_{1} = 3

and thickness

\frac{d}{3}

while the other one has dielectric constant

K_{2} = 6

and thickness

\frac{2 d}{3}

. Capacitance of the capacitor is now :

A parallel plate capacitor with air between the plates has a capacitance of $9 p F$ . The separation between its plates is $d$ . The space between the plate is now partially filled with two dielectrics. One of the dielectric has dielectric constant $K_{1} = 3$ and thickness $\frac{d}{3}$ while the other one has dielectric constant $K_{2} = 6$ and thickness $\frac{d}{3}$ . The remaining $\frac{d}{3}$ space is filled with air. The new capacitance is :