Electrostatics Question 765
Question: If the electrostatic potential were given by $ \phi ={\phi _{0}}(x^{2}+y^{2}+z^{2}), $ where is constant, then the charge density giving rise to the above potential would be.
Options:
A) 0
B) $ -6{\phi _{0}}{\varepsilon _{0}} $
C) $ -2{\phi _{0}}{\varepsilon _{0}} $
D) $ -\frac{6{\phi _{0}}}{{\varepsilon _{0}}} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ E=-\nabla \phi =-{\phi _{0}}2[x\hat{i}+y\hat{i}+z\hat{x}] $
$ ={\varepsilon _{0}}\nabla .E=-2{\varepsilon _{0}}{\phi _{0}}\nabla .(x\hat{i}+y\hat{i}+z\hat{x}) $
$ n=-6{\phi _{0}}{\varepsilon _{0}} $
