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The electric potential at a point $\;(x , y)\;$ in the x- y plane is given by $\;V=-k\;x\;y\;$. The field intensity at a distance r in this plane , from the origin is proportional to

$(a)\;r^2\qquad(b)\;r\qquad(c)\;\large\frac{1}{r}\qquad(d)\;\large\frac{1}{r^2}$

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Answer : (b) r
Explanation :
$E_{x}=-\large\frac{\partial V}{\partial x}=ky$
$E_{y}=-\large\frac{\partial V}{\partial y}=kx$
$\overrightarrow{E}=E_{x} \hat{i} + E_{y} \hat{j}$
$\overrightarrow{E}=k_{y} \hat{i} +k_{x} \hat {j}$
$|\overrightarrow{E}|=\sqrt{k^2 y^2 + k^2 x^2}$
$|\overrightarrow{E}|= k\; \sqrt{x^2+y^2}$
$=k\;r$
Thus $\; |\overrightarrow{E}| \propto r $
answered Feb 19, 2014 by yamini.v
 

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