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>>Electrostatic Potential and Capacitance
>>Effects of Dielectrics in Capacitors
>> $A parallel plate condenser is filled wit$

Question

A parallel plate condenser is filled with two dielectrics as shown in figure. Area of each pate is $A m^{2}$ and the separation is $d$ metre. The dielectric constants are $K_{1}$ and $K_{2}$ respectively. Its capacitance (in farad) will be :

466798

A

$\frac{2 \in _{0} A}{d} (\frac{K _{1} + K _{2}}{K _{1} K _{2}})$

B

$\frac{2 \in _{0} A}{d} (\frac{K _{1} K _{2}}{K _{1} + K _{2}})$

C

$\frac{\in _{0} A}{d} (\frac{K _{1} + K _{2}}{2 K _{1} K _{2}})$

D

$\frac{\in _{0} A K _{1} K _{2}}{2 ( d _{2} K _{1} + d _{1} K _{2} )}$

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Updated on : 2022-09-05

Solution

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Correct option is B)

Capacitance of first half of the capacitor of width $d / 2 = C_{1} = \frac{K _{1} A ϵ _{0}}{d / 2} = \frac{2 K _{1} A ϵ _{0}}{d}$
Similarly the capacitance of other half $= C_{2} = \frac{2 K _{2} A ϵ _{0}}{d}$
These two capacitance are connected in series.
Thus, the net capacitance of the combination is, $C_{e f f} = \frac{C _{1} C _{2}}{C _{1} + C _{2}} = \frac{2 A ϵ _{0}}{d} (\frac{K _{1} K _{2}}{K _{1} + K _{2}})$

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