The plates of a parallel plate capacitor are charged up to 100 V. Now, after removing the battery, a 2 mm thick plate is inserted between the plates . Then, to maintain the same potential difference, the distance between the capacitor plates is increased by 1.6 mm. The dielectric constant of the plate is A) 5 B) 1.25 C) 4 D) 2.5

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Gregory Ferguson · Accepted Answer

The correct answer is A 5As the battery is disconnected, charge q will remain the same.It is given that final potential V is the same.From q=CV , we can say that the final capacitance should  also be same as the initial capacitance i.e.,      C    1    =      C    2  ....(1)If the area of the plates is A and the intial separation between them is d then the capacitance will be      C    1    =                              ε                0            A        d  ....(2)As previously stated, the capacitance of a partially filled dielectric slab of thickness t and dielectric constant K is given by      C    1    =                              ε                0            A        d  ....(3)​Given,t=2 mm and d′=d+1.6From equation (1), (2) and (3), we have                              ε                0            A        d    =                              ε                0            A              (      d      +      1.6      )      −      2      +              2        K            On comaparing the both sides term we get,  1.6  −  2  +      2    K    =  0  ⇒      2    K    =  0.4Thus, K=5Therefore, option (a) is the correct answer.Alternate Solution:Let, the initial voltage across the capacitor is       V    0   and  the initial Electric field be       E    0    ⇒      V    0    =      E    0    d....(1)Let dielectric constant be K. So, the new voltage,  V      ′    =      E    0    (  d  −  t  +      t    K    )Given, t=2 mmTo increase the voltage from V′ to       V    0  , separation between the plates is increased by 1.6 mm. Thus,      V    0    =      E    0    (  d  −  t  +      t    K    +  1.6  )....(2)Comparing the equations (1) and (2), we get  −  t  +      t    K    +  1.6  =  0  ⇒  −  2  +      2    K    +  1.6  =  0⇒K=5Hence, option (a) is correct.Key concept:1. The electric field inside the dielectric of an isolated charged capacitor is reduced by the dielectric constant factor.2. The electric field of a parallel plate capacitor can be assumed to be uniform if the distance between the plates is small.

The plates of a parallel plate capacitor are charged up to 100V. Now, after removing the battery, a 2 mm thick plate is inserted between the plates . Then, to maintain the same potential difference, the distance between the capacitor plates is increased by 1.6 mm.

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