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A parallel plate capacitor with square plates is filled with four dielectrics constant K1 , K2, K3 , K4 arranged as shown in the figure. The effective dielectric constant K will be:

  • Option 1)

    K = \frac{K_1K_2}{K_1+K_2}+\frac{K_3K_4}{K_3+K_4}

  • Option 2)

    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 )}

  • Option 3)

    K = \frac{\left ( K_{1}+K_{2} \right ) \left ( K_{3}+K_{4} \right )}{K_{1}+K_{2}+K_{3}+K_{4}}

  • Option 4)

    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 )}

Answers (1)

best_answer

 

If K filled between the plates -

{C}'=K\frac{\epsilon _{0}A}{d}={C}'=Ck

 

 

- wherein

C\propto A

C\propto\frac{1}{d}

 

C_{1}= \frac{K_{1}\epsilon _{0}\frac{L}{2}\times L}{d/2}

C_{2}= \frac{K_{2}\epsilon _{0}\frac{L}{2}\times L}{d/2}

C_{3}= \frac{K_{3}\epsilon _{0}\frac{L}{2}\times L}{d/2}

C_{4}= \frac{K_{4}\epsilon _{0}\frac{L}{2}\times L}{d/2}

C_{g}=\frac{C_{1}C_{2}}{C_{1}+C_{2}}+\frac{C_{3}C_{4}}{C_{3}+C_{4}}

K = \frac{K_1K_2}{K_1+K_2}+\frac{K_3K_4}{K_3+K_4}

 

 


Option 1)

K = \frac{K_1K_2}{K_1+K_2}+\frac{K_3K_4}{K_3+K_4}

Option 2)

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 )}

Option 3)

K = \frac{\left ( K_{1}+K_{2} \right ) \left ( K_{3}+K_{4} \right )}{K_{1}+K_{2}+K_{3}+K_{4}}

Option 4)

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 )}

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