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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 5

  • question_answer
    The plates of a parallel plate capacitor are charged up to 100 V. 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 mm. The dielectric constant of the plate is

    A) 5         

    B) 1.25  

    C) 4                     

    D) 2.5

    Correct Answer: A

    Solution :

    [a] : Here, t = 2 mm, x = 1.6 mm, K = ?      As potential difference remains the same, capacity must remain the same                        \[\therefore \]\[\frac{{{\varepsilon }_{0}}A}{d-t+\frac{t}{k}}=\frac{{{\varepsilon }_{0}}A}{d-1.6}\]where d is the separation between the plates initially. \[x=t\left( 1-\frac{1}{K} \right)\Rightarrow 1.6=2\left( 1-\frac{1}{K} \right)\]which gives K=5.


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