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The electric potential V at any point (x,y,z) in space is given by $V= gx^2\; volt$. Electric field at $ (2,0,5) m $ in $Vm^{-1}$ is :

$(A)\;36, along\;+ve \; x-axis \\ (B)\;36, along\;-ve \; x-axis \\ (C)\; 90, along\;+ve \; z-axis \\ (D)\;90, along\;-ve \; z-axis $

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$E= \large\frac{-dv}{dr}$
$\quad= \large\frac{-d (x^2)}{dx}$
$\quad= -18 x$
=> $E= -18 (2) =-36 Vm^{-1}$
Hence B is the correct answer.
answered Jan 8, 2014 by meena.p

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