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CBSE Class 12 Physics 2016 Delhi Set 1 Paper

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Question : 15 of 26
Marks: +1, -0
Two parallel plate capacitors XX and YY have the same area of plates and same separation between them. XX has air between the plates while bo{Y}bo\{Y\} contains a dielectric medium of εr=4\varepsilon_{r}=4.
(i) Calculate capacitance of each capacitor if equivalent capacitance of the combination is 4  μF4\;\mu\mathrm{F}.
(ii) Calculate the potential difference between the plates of XX and YY.
(iii) Estimate the ratio of electrostatic energy stored in XX and YY.
Solution:  
(i) Calculation of capacitance of each capacitor
(ii) Calculation of potential difference
(iii) Estimation of ratio of electrostatic energy
(i) Let CX=CC_X=C
Cγ=4CC_\gamma=4C (as it has a dielectric medium of εγ\varepsilon_\gamma =4=4 )
For series combination of two capacitors
  1C  =  1CX+  1CY\;\frac{1}{C}\;=\;\frac{1}{C_X}+\;\frac{1}{C_Y}
⇒      14  μF  =  1C+  14C\Rightarrow\;\;\;\frac{1}{4\;\mu\mathrm{F}}\;=\;\frac{1}{C}+\;\frac{1}{4C}
  14  μF  =  54C\;\frac{1}{4\;\mu\mathrm{F}}\;=\;\frac{5}{4C}
⇒    C  =5  μF\Rightarrow\;\;C\;=5\;\mu\mathrm{F}
   Hence       CX  =5  μF\;\text{ Hence }\;\;\;C_X\;=5\;\mu\mathrm{F}
CY  =20  μFC_Y\;=20\;\mu\mathrm{F}
Hence CX=5  μFC_X=5\;\mu\mathrm{F}
Cγ=20  μFC_\gamma=20\;\mu\mathrm{F}
     (ii) Total charge   Q=CV\;\;\text{ (ii) Total charge }\; Q=C V
  =4  μF×15  V=60  μC\;=4\;\mu\mathrm{F} \times 15\;\mathrm{V}=60\;\mu\mathrm{C}
  VX=  QCX=  60  μC5  μF=12  V\;V_X=\;\frac{Q}{C_X}=\;\frac{60\;\mu\mathrm{C}}{5\;\mu\mathrm{F}}=12\;\mathrm{V}
  VY=  QCY=  60  μC20  μF=3  V\;V_Y=\;\frac{Q}{C_Y}=\;\frac{60\;\mu\mathrm{C}}{20\;\mu\mathrm{F}}=3\;\mathrm{V}
     (iii)   ExEy=  Q22CXQ22CY\;\;\text{ (iii) }\; \frac{E_x}{E_y}=\;\frac{\frac{Q^2}{2C_X}}{\frac{Q^2}{2C_Y}} =  CYCX=  205=4:1=\;\frac{C_Y}{C_X}=\;\frac{20}{5}=4:1
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