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  • Need clarity, kindly explain! - Electrostatics - JEE Main-7

A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric
constant K=\frac{5}{3}   is inserted between the plates, the magnitude of the induced charge will be :

  • Option 1)

    0.9 n C
     

  • Option 2)

    1.2 n C
     

  • Option 3)

    0.3 n C
     

  • Option 4)

    2.4 n C

 

Answers (2)

best_answer

Polarisation=Q(1-\frac{1}{K})

                              =90*10^{-12}*20*\frac{5}{3}(1-\frac{3}{5})

                             =1200*10^{-12}C

                             =1.2 nc

 

Polarisation of Dielectric slab -

It is the process inducing equal and opposite charge on the two faces of the dielectric.

- wherein

 

 


Option 1)

0.9 n C
 

This is incorrect

Option 2)

1.2 n C
 

This is correct

Option 3)

0.3 n C
 

This is incorrect

Option 4)

2.4 n C

This is incorrect

Posted by

prateek

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