The electric potential at a point (x, y) in the x - y plane is given by V = - kxy. The field intensity at a distance r from the origin varies as

r^{2}

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r

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\frac{1}{r}

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\frac{1}{r^{2}}

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Solution

The correct option is B $r$

$E_{x} = - \frac{d V}{d x} = - k y; E_{y} = - \frac{d V}{d y} = - k x$
$\Rightarrow E = \sqrt{E_{x}^{2} + E_{y}^{2}} = k \sqrt{x^{2} + y^{2}} = k r \Rightarrow E \propto r$

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The electric potential at a point (x, y) in the x - y plane is given by V = - kxy. The field intensity at a distance r from the origin varies as

The correct option is B r Ex=−dVdx=−ky; Ey=−dVdy=−kx ⇒E=√E2x+E2y=k√x2+y2=kr⇒E∝r

The correct option is B $r$

$E_{x} = - \frac{d V}{d x} = - k y; E_{y} = - \frac{d V}{d y} = - k x$
$\Rightarrow E = \sqrt{E_{x}^{2} + E_{y}^{2}} = k \sqrt{x^{2} + y^{2}} = k r \Rightarrow E \propto r$