Question:

A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants $K_1, K_2, K_3, K_4$ arranged as shown in the figure. The effective dielectric constant K will be :

Updated On: Sep 27, 2024
  • $K = \frac{\left(K_{1} +K_{2} \right)\left(K_{3}+ K_{4}\right)}{2\left(K_{1} + K_{2}+K_{3} +K_{4}\right)} $
  • $K = \frac{\left(K_{1} +K_{2} \right)\left(K_{3}+ K_{4}\right)}{\left(K_{1} + K_{2}+K_{3} +K_{4}\right)} $
  • $K = \frac{\left(K_{1} +K_{4} \right)\left(K_{2}+ K_{3}\right)}{2\left(K_{1} + K_{2}+K_{3} +K_{4}\right)} $
  • $K = \frac{\left(K_{1} +K_{3} \right)\left(K_{2}+ K_{4}\right)}{K_{1} + K_{2}+K_{3} +K_{4}} $
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The Correct Option is D

Solution and Explanation

$C_{12} = \frac{C_{1} C_{2}}{C_{1 }+ C_{2}} = \frac{\frac{k_{1} \in_{0} \frac{L}{2} \times L }{d/2} . \frac{k_{2} \left[\in_{0} \frac{L}{2} \times L\right]}{d/2}}{\left(k_{1} +k_{2}\right) \left[\frac{\in_{0} . \frac{L}{2} \times L}{d/2}\right]} $
$ C_{12} = \frac{k_{1}k_{2}}{k_{ 1} +k_{2}} \frac{\in_{0} L^{2}}{d} $
in the same way we get, $ C_{34} = \frac{k_{3}k_{4} }{k_{3} +k_{4}} \frac{ \in_{0} L^{2}}{d} $
$ \therefore C_{eq} = C_{12} + C_{34} = \left[\frac{k_{1}k_{2}}{k_{1} + k_{2}} + \frac{k_{3}k_{4}}{k_{3} + k_{4}} \right] \frac{\in_{0} L^{2}}{d} $
....(i)
Now if $ k_{eq} = k , C_{eq} = \frac{k \in_{0} L^{2}}{d} $ .....(ii)
on comparing equation (i) to equation (ii), we get
$ k_{eq} = \frac{k_{1}k_{2} \left(k_{3} +k_{4}\right)+k_{3} k_{4} \left(k_{1 }+k_{2}\right)}{\left(k_{1} +k_{2} \right)\left(k_{3}+ k_{4}\right)} $
This does not match with any of the options so probably they have assumed the wrong combination
$ C_{13} = \frac{k_{1} \in_{0} L \frac{L}{2}}{d/2} + k_{3} \in_{0} \frac{L. \frac{L}{2}}{d/2} $
$ = \left(k_{1} + k_{3}\right) \frac{\in_{0} L^{2}}{d} $
$ C_{24} = \left(k_{2 } + k_{4}\right) \frac{\in_{0} L^{2}}{d} $
$ C_{eq} = \frac{C_{13} C_{24}}{C_{13} C_{24}} = \frac{\left(k_{1} + k_{3} \right)\left(k_{2} +k_{4}\right) }{\left(k_{1} + k_{2} + k_{3} + k_{4}\right)} \frac{\in _{0} L^{2}}{d} $
$ = \frac{k \in_{0}L^{2}}{d} $
$ k = \frac{\left(k_{1} +k_{3}\right)\left(k_{2} +k_{4}\right)}{\left(k_{1} +k_{2 }+ k_{3} + k_{4}\right) }$
However this is one of the four options. It must be a "Bonus" logically but of the given options probably they might go with (4)
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Concepts Used:

Electrostatic Potential and Capacitance

Electrostatic Potential

The potential of a point is defined as the work done per unit charge that results in bringing a charge from infinity to a certain point.

Some major things that we should know about electric potential:

  • They are denoted by V and are a scalar quantity.
  • It is measured in volts.

Capacitance

The ability of a capacitor of holding the energy in form of an electric charge is defined as capacitance. Similarly, we can also say that capacitance is the storing ability of capacitors, and the unit in which they are measured is “farads”.

Read More: Electrostatic Potential and Capacitance

The capacitor is in Series and in Parallel as defined below;

In Series

Both the Capacitors C1 and C2 can easily get connected in series. When the capacitors are connected in series then the total capacitance that is Ctotal is less than any one of the capacitor’s capacitance.

In Parallel

Both Capacitor C1 and C2 are connected in parallel. When the capacitors are connected parallelly then the total capacitance that is Ctotal is any one of the capacitor’s capacitance.