JEE Main & Advanced Sample Paper JEE Main Sample Paper-20

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

    A)  \[-20\,V/m\]                  

    B)  \[6\,V/m\]

    C)  \[11\,V/m\]                   

    D)  \[+20\,V/m\]

    Correct Answer: A

    Solution :

     \[E=\frac{-dV}{dx}V=5{{x}^{2}}+10x-9,\] \[E=\frac{-d}{dx}(5{{x}^{2}}+10x-9)=-(10x+10)\] On putting value of x in it we get, \[E=-(10\times 1+10)=-20V/m\]

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