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

  • question_answer
    In a region, the potential is represented by \[V(x,\,y,\,z)=6x-8xy-8y+6yz,\] where V is in volts and x, y, z are in metres. The electric force experienced by a charge of 2 coulomb situated at point \[(1,\text{ }1,\text{ }1)\] is           [NEET 2014]

    A) \[6\sqrt{5}N\]

    B) \[30\,N\]

    C) \[24\,N\]

    D) \[4\sqrt{35}N\]

    Correct Answer: D

    Solution :

    [d] We know
    \[F=qE\]                                                …(i)
    \[E=-\frac{dv}{dr}\]
                            \[\left. \begin{align}   & {{E}_{x}}\frac{\delta V}{\delta x}=6-8y \\  & {{E}_{y}}=\frac{\delta V}{\delta y}=-8x-8+6z \\ \end{align} \right\}\]       …(ii)
                            \[{{E}_{z}}=6y\]
    Above values of \[{{E}_{x}},{{E}_{y}}\], and \[{{E}_{z}}\] at \[(1,\text{ }1,\text{ }1)\] are   
                            \[{{E}_{x}}=6-8\times (1)=-2\] 
                            \[{{E}_{y}}=-8(1)-8+6(1)=-10\]
    \[{{E}_{z}}=6\times 1=6\]
    So,       \[{{E}_{net}}=\sqrt{{{(-2)}^{2}}+{{(10)}^{2}}+{{(6)}^{2}}}\]
    \[=\sqrt{4+100+36}=\sqrt{40}\Rightarrow \sqrt{35\times 4}\]
    \[=2\sqrt{35}N/C\]
    So,       \[F=q\,{{E}_{net}}=2(2\sqrt{35})=4\sqrt{35}N\]


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