Electrostatics Question 776

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

Options:

A) $ r^{2} $

B) $ r $

C) $ \frac{1}{r} $

D) $ \frac{1}{r^{2}} $

Show Answer

Answer:

Correct Answer: B

Solution:

[b] $ \vec{E}=\frac{\partial v}{\partial x}\hat{i}+\frac{\partial v}{\partial y}\hat{j} $

$ \therefore |\vec{E}|=k( \sqrt{x^{2}+y^{2}} )=kr $ Given $ v=-kxy\text{ }E\propto r $

$ \therefore \vec{E}=ky\hat{i}+ky\hat{j} $



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