Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Coulomb's law and Maxwell's equations should be consistant as one can be derived from the other.

Say we have a point charge with such a charge that $-kq=1$, meaning that at any point the electric field will have a magnitude of

$$|E|=\frac{1}{r^2}$$

where $r$ is the distance from the origin (were we place our charge), and the vectors point towards the origin at all point. This would be equivalent to the following in cartesian co-ordinates:

$$E=-\hat{x}\frac{x}{(x^2+y^2+z^2)^{3/2}}-\hat{y}\frac{y}{(x^2+y^2+z^2)^{3/2}}-\hat{z}\frac{z}{(x^2+y^2+z^2)^{3/2}}$$ We can verify that

$$|E|=\frac{1}{\sqrt{x^2+y^2+z^2}}$$

Gauss's law in its differential form allows us to calculate a charge distribution that would give rise to such a electric field using the divergence operator:

$$\nabla \cdot E=-\frac{1}{x^2+y^2+z^2}$$

from wolfram alpha: one, two

Which absolutely doesn't make sense to me! Intuitively I would think it would be zero everywhere except (0,0,0). Or at least not to go of to infinity at any point.

Could somebody please explain what's going on?

share|improve this question
add comment

1 Answer 1

Actually if you calculate $\boldsymbol{\nabla} \cdot \mathbf{E}$, you get zero except at the origin, where you get infinity. So you can do it more precisely and obtain a delta function. I suspect an error in your calculation.

share|improve this answer
1  
The error is he says $|\mathbf{E}|$ goes like $1/r$. –  NowIGetToLearnWhatAHeadIs Jan 31 at 0:36
    
In other words, he should have done this calculation instead. Adding up gives a denominator of $2x^2 + 2y^2 + 2z^2 - (y^2 + z^2) - (x^2 + z^2) - (x^2 + y^2) = 0$ –  lionelbrits Jan 31 at 17:17
1  
You meant numerator, but yes I agree with you. –  NowIGetToLearnWhatAHeadIs Jan 31 at 17:26
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.