4
$\begingroup$

Are the charge distributions $$\rho(\vec{r})=\frac{Q}{2\pi R^2}\delta(r-R)\delta(\vartheta-\pi/2)$$ and $$\rho(\vec{r})=\frac{Q}{2\pi r^2\sin(\vartheta)}\delta(r-R)\delta(\vartheta-\pi/2)$$ of a charged circle the same? I would say yes because integrating over them gives the same result but is this in general true?

$\endgroup$

2 Answers 2

4
$\begingroup$

Yes, since $\delta(r-R)\delta(\vartheta-\pi/2)$ is zero everywhere except $\left(r,\vartheta\right)=\left(R,\theta/2\right)$, we can replace $$ \frac{Q}{2\pi r^2\sin(\vartheta)} $$ with its value at $\left(r,\vartheta\right)=\left(R,\theta/2\right)$: $$ \frac{Q}{2\pi R^2} $$ If you have a copy of Griffiths E&M he discusses this property of the Dirac delta function (section 1.5.2, equation 1.88, at least in the Third Edition).

$\endgroup$
4
$\begingroup$

Two distributions are defined as being equal if, when integrated with respect to an arbitrary test function, they always yield the same result. In other words, $D_1(r,\theta)$ and $D_2 (r, \theta)$ are equal if for all test functions $f(r, \theta)$, we have $$ \int D_1(r,\theta) f(r, \theta) = \int D_2(r,\theta) f(r, \theta). $$ If you have a copy of Griffiths, this is discussed briefly in Section 1.5.2 (Equation 1.93 et seq.)

In your case, it is straightforward to evaluate both integrals and show that both are equal to $\frac{Q}{2 \pi R} f(R, \pi/2)$ regardless of your choice of $f$. Thus, the two distributions are equal.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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