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

  • question_answer
    The electric field in a region is given by \[\overrightarrow{E}=\left( Ax+B \right)\hat{i},\] where E is in \[N{{C}^{-1}}\] and x is in metres. The values of constants are A = 20 SI unit and B = 10 SI unit. If the potential at x = 1 is \[{{V}_{1}}\]and that at x = -5 is \[{{V}_{2}}\], then \[{{V}_{1}}-{{V}_{2}}\] is :- [JEE Main 8-4-2019 Afternoon]

    A) - 48 V

    B) - 520 V

    C) 180 V

    D) 320 V

    Correct Answer: C

    Solution :

    [c] \[\vec{E}=(20x+10)\hat{i}\]
                \[{{V}_{1}}-{{V}_{2}}-\int\limits_{-5}^{1}{(20x+10)dx}\]
                \[{{V}_{1}}-{{V}_{2}}=-(10{{x}^{2}}+10x)_{-5}^{1}\]
                \[{{V}_{1}}-{{V}_{2}}=10(25-5-1-1)\]
                \[{{V}_{1}}-{{V}_{2}}=180V\]


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