A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage (U ) as ε = αU where α = 2V^{-1} .A similar capacitor with no dielectric is charged to U0 = 78 V. It is then connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors.

NCERT Class XII
Physics - Exemplar Problems

Chapter 2. Potential and Capacitance

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2015-06-10T15:58:02+05:30

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Dielectric medium:  relative permittivity ε = α U   
           so permittivity :  ε ε₀ = α U ε₀
       where  U = applied voltage  and  α = 2 / volt

 Capacitance C₁ = Q / V = ε ε₀ A / d = 2 U ε₀ A / d
 Capacitance C₂ = ε₀ A / d

               C₁ = C₂ * 2 U

  Given,  V₂ = U₀ = 78 Volts.
    
Initially, the Capacitor C1 with the dielectric medium is not charged.   Hence,
         Q₁ = V₁ * C₁  = 0
The charge on capacitor C₂ :   Q₂ = C₂ V₂  = C₂ * 78  Coulombs

When the two capacitances are connected in parallel, the charges are distributed among the both.  The equivalent capacitance is C = C₁ + C₂.  The voltage across both capacitors  will be same.

      C = C₂ (1 + 2 U) 
      Q = charge on both capacitors = Q₂ = 78 C₂

Final voltage across both :   U = Q / C  =  78 / (1 + 2 U)
       =>  U + 2 U² - 78 = 0
       =>  U = [ -1 + - √(1 + 624) ] / 4
       =>  U = 6 Volts.

So the capacitance of C₁ = 12 C₂  Farads    and potential difference is 6 Volts.

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