The electric potential at a point $(x, y)$ is given by: $V=-K x y .$ The electric field intensity a distance…
The electric potential at a point $(x, y)$ is given by: $V=-K x y .$ The electric field intensity a distance $r$ from the origin varies as
- $r^{2}$
- r
- $2 r$
- $2 r^{2}$
Solution
$\vec{E}=\left[\frac{\partial V}{\partial x} \hat{i}+\frac{\partial V}{\partial y} \hat{j}\right]=K(\hat{v}+x \hat{j}) \therefore E=\sqrt{E_{x}^{2}+E_{y}^{2}}=\sqrt{(K y)^{2}+(K x)^{2}}=\operatorname{Kr}$
i.e., $E \propto r$
$\therefore$ (b)
Asked in: JEE Mains - Electrostatics - Test 3
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