JEE Main & Advanced Physics Electrostatics & Capacitance JEE PYQ-Electrostatics and Capacitance

  • question_answer
    A parallel plate capacitor has\[1\mu F\] capacitance. One of its two plates is given\[+2\mu C\]charge and the other plate, \[+4\mu C\] charge. The potential difference developed across the capacitor is:- [JEE Main 8-4-2019 Afternoon]

    A) 5V

    B) 2V

    C) 3V

    D) 1V

    Correct Answer: D

    Solution :

    [d]
    Charges at inner plates are \[1\mu C\]and \[1\mu C\]
    \[\therefore \]Potential difference across capacitor
    \[=\frac{q}{c}=\frac{1\mu C}{1\mu F}=\frac{1\times {{10}^{-6}}C}{1\times {{10}^{-6}}Farad}=1V\]


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