The electric potential at a point $(x, y)$ in the $x-y$ plane is given $\mathrm{by}…
The electric potential at a point $(x, y)$ in the $x-y$ plane is given $\mathrm{by} \mathrm{V}=-\mathrm{kxy}$. The field intensity at a distance $\mathrm{r}$ from the origin varies as
\(\mathrm{r}^{2}\)
\(\frac{1}{\mathrm{r}^{2}}\)
\(1 / \mathrm{r}\)
\(r\)
Solution
Let the \(x-y\) coordinates of the point at distance \(r\) from the origin be given as \(x=r \cos \theta, y=r \sin \theta\)
Potential is given as \(\mathrm{V}=-\mathrm{Kxy}\)
Now, Electric field intensity is \(\vec{E}=(-\mathrm{dV} / \mathrm{dx}) \vec{i}+(-\mathrm{dV} / \mathrm{dy}) \vec{j}=\mathrm{K}(\mathrm{y} \vec{i}+\mathrm{x} \vec{j})=\)
\(\mathrm{K}(r \sin \theta \vec{i}+\mathrm{r} \cos \theta \vec{j})=\mathrm{Kr}(\sin \theta \vec{i}+\cos \theta \vec{j}) \propto \mathrm{r}\)
So electric field potential is proportional to \(\mathrm{r}\).