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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Electric Field and Potential

  • question_answer
    The electric potential \[V\] is given as a function of distance \[x\] (metre) by \[V=(5{{x}^{2}}+10x-9)\,volt\]. Value of electric field at \[x=1\] is                                         [MP PET 1999]

    A)                    \[20\,V/m\]

    B)                                      \[6\,V/m\]

    C)                    \[11\,V/m\]

    D)                                      \[-23\,V/m\]

    Correct Answer: A

    Solution :

                         \[E=-\frac{dV}{dx}=-\frac{d}{dx}(5{{x}^{2}}+10x-9)=-10x-10\] \ \[{{}_{x=1}}=-10\times 1-10=-\,20\,V/m\]


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