Question

3. The electric potential at a point (x, y) in the xy-
plane is given by:
V=-Kxy
The electric field intensity at a distancer from the
origin varies as:
(a)r² (b) (c) 21 (d) 2r²

Plzz solve this​

Answer 1

$\bold {\huge {Your ~answer :-}}$

$\bf\huge\boxed{r}$

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EXPLAINATION ➣

Let the x—y coordinates of the point at distance r from the origin be given as

x = r cosθ , y = r sinθ

Potential is given as V = - Kxy

Now, $\green{\texttt{Electric field intensity }}$ is

E = (-dV/dx) i + (-dV/dy) j

E = K(yi + xj)

E = K(r sinθ i + r cosθ j)

E = Kr (sinθ i + cosθ j) ∝ r

So electric field potential is $\green{\texttt{proportional }}$ to r.

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Note ⇒ kindly revise the vector sum and products once

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