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Question

A capacitor of capacitance 1 blank mu F is filled with two dielectrics of dielectric constant 4 and 6. What is the new capacitance?

  1. 10 mu F    
  2. 5 mu F    
  3. 4 mu F    
  4. 7 mu F    

The correct answer is: 5 mu F


    Initially, the capacitance of capacitor

    C equals fraction numerator epsilon subscript 0 end subscript A over denominator d end fraction
    therefore blank fraction numerator epsilon subscript 0 end subscript A over denominator d end fraction equals 1 mu F …(i)
    When it is filled with dielectric of dielectric constant K subscript 1 end subscript a n d blank K subscript 2 end subscript as shown, then there are two capacitors connected is parallel. So,
    C to the power of ´ ´ end exponent equals fraction numerator K subscript 1 end subscript epsilon subscript 0 end subscript open parentheses fraction numerator A over denominator 2 end fraction close parentheses over denominator d end fraction plus fraction numerator K subscript 2 end subscript epsilon subscript 0 end subscript open parentheses fraction numerator A over denominator 2 end fraction close parentheses over denominator d end fraction
    (as area becomes half)
    C ´ ´ blank equals fraction numerator 4 epsilon subscript 0 end subscript A over denominator 2 d end fraction plus fraction numerator 6 epsilon subscript 0 end subscript A over denominator 2 d end fraction equals fraction numerator 2 epsilon subscript 0 end subscript A over denominator d end fraction plus fraction numerator 3 epsilon subscript 0 end subscript A over denominator d end fraction
    Using Eq. (i), we obtain
    C to the power of ´ ´ end exponent equals blank 2 cross times 1 plus 3 cross times 1 equals 5 blank mu F

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