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

Question

Dr. Abubacker · Accepted Answer

(b) Co-ordinates of the point are (x, y)
$	herefore\ \ $Distance of point from origin,
$r = \sqrt{x^{2} + y^{2},\ }\ V = - kxy$
$E_{x} = - \frac{dV}{dx} = - \frac{d}{dx}( - kxy) = ky$
$E_{y} = - \frac{dV}{dy} = ( - kxy) = kx$
$	herefore\ E = \sqrt{E_{x}^{2} + E_{y}^{2}} = k\sqrt{y^{2} + x^{2}} = kr$
$	herefore\ E \propto r$

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

Explanation

Reviewed by Dr. Abubacker

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