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} $
