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}} $



NCERT Chapter Video Solution

Dual Pane