12th Class Physics Electrostatics & Capacitance Question Bank Case Based (MCQs) - Electrostatic Potential and Capacitance

  • question_answer
    Direction: Q.31 to Q.35
    The simplest and the most widely used capacitor is the parallel plate capacitor. It consists of two large plane parallel conducting plates, separated by a small distance.
    In the outer regions above the upper plate and below the lower plate, the electric fields due to the two charged plates cancel out. The net field is zero.
    In the inner region between the two capacitor plates, the electric fields due to the two charged plates add up. The net field is \[\frac{\sigma }{{{\varepsilon }_{0}}}\].
    For a uniform electric field, potential difference between the plates = Electric field \[\times \] distance between the plates.
    Capacitance of the parallel plate capacitor is, the charge required to supplied to either of the conductors of the capacitor so as to increase the potential difference between, then by unit amount.
    Read the given passage carefully and give the answer of the following questions.
    A parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on charge, potential and capacitance respectively are:

    A) increases, decreases, decreases

    B) constant, increases, decreases

    C) constant, decreases, decreases

    D) constant, decreases, increases

    Correct Answer: B

    Solution :

    (b) constant, increases, decreases                         As the capacitor is isolated after charging, charge Q on it remains constant. Plate separation d increases, capacitance decreases as \[C=\frac{{{\varepsilon }_{0}}A}{d}\]and hence, potential increases as \[V=\frac{Q}{C}.\]


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