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JCECE Engineering JCECE Engineering Solved Paper-2006

  • question_answer
    The electric potential \[V\] is 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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