JCECE Medical JCECE Medical Solved Paper-2006

  • question_answer
    The electric potential Vis given as a function of distance \[x\](metre) by \[V=(5{{x}^{2}}+10x-4)V.\]Value of electric field at \[x=1\,m\]is:

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

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

    C)  \[6\,V/m\]

    D)  \[-20\,V/m\]

    Correct Answer: D

    Solution :

     Key Idea: \[E=-\frac{dV}{dx}\] Electric field \[E\] is given by \[E=-\frac{dV}{dx}\] Given, \[V=5{{x}^{2}}+10x-4\] \[\therefore \] \[E=\frac{dV}{dx}=10x+10\] At \[x=1,E=-(10\,\,\times \,\,1\,+\,10)=-20\,V/m\]


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