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Electrostatic Potential and Capacitance
Electric potential at any point is $V=-10 x +6y +\sqrt {60 }z$, then magnitude of electric field is :
$(A)\;6 \sqrt 2 \\ (B)\;8 \sqrt 2 \\ (C)\;14 \\ (D)\;10 \sqrt 2 $
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Jan 3, 2014
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meena.p
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Aug 22, 2014
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1 Answer
$V= -10 x +6y +\sqrt {60}z$
$E_x=\large\frac{-dV}{dx}$$=10$
$E_y=\large\frac{-dV}{dy}$$=-6$
$E_z=\large\frac{-dV}{dz}$$=- \sqrt {60}$
$|E|=\sqrt {E_x^2 +E_y^2+E_z^2}$
$\qquad= {100 +36+60}$
$\qquad=14$
Hence C is the correct answer.
answered
Jan 3, 2014
by
meena.p
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