12th Class Physics Electrostatics & Capacitance Question Bank MCQs - Electrostatic Potential and Capacitance

  • question_answer
    The capacitance of a parallel plate capacitor is 10 \[\mu \operatorname{F}\]. When a dielectric plate is introduced in between the plates, its potential becomes 1/4th of its original value. What is the value of the dielectric constant of the plate introduced?

    A) 4

    B) 40

    C) 2.5

    D) none of the above

    Correct Answer: A

    Solution :

    Option [a] is correct
    Explanation: C' = KC (where K is the dielectric constant).
    V=Q/C
    V'=Q/C'   
    V' = V/4 = Q/C' a Q/KC = V/K
    \[\therefore \]  K=4


You need to login to perform this action.
You will be redirected in 3 sec spinner