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NEET AIPMT SOLVED PAPER SCREENING 2006

  • question_answer
                    A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates:                                                                                    

    A)                  decreases     

    B)                  does not change

    C)                  becomes zero   

    D)                  increases

    Correct Answer: D

    Solution :

                    Key Idea: Charge remains constant after charging.                 If the battery is removed after charging then the charge stored in the capacitor remains constant.                 q = constant                 Change in capacitance                 \[C'=\frac{{{\varepsilon }_{0}}\,A}{d'}\] \[Asd'>d\] hence,  \[C'<C\]                 Hence, potential difference between the plates                 or                            \[V'\,\,\propto \,\,\frac{1}{C'}\]                 As capacitance decreases, so potential difference increases.                 Note:    If the battery remains connected, the charge stores increases. Also the potential difference V becomes constant.


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