Why Is a Negative Sign Used in the Electric Potential of a Charged Sphere?

[Shinobii]

Hello,

The electric potential is defined as:

$$
\phi(\vec{r}) = \int d^3r' \frac{\rho(\vec{r}')}{|\vec{r} - \vec{r}'|}.
$$

My question is, for solving for the potential inside of a charged solid sphere (constant charge density) by using the above equation I get,

$$
\frac{Q}{2R} \bigg( \frac{r^2}{R^2} - 3 \bigg).
$$

When the result is actually the above multiplied by a minus sign. Is this because when we take the reference point to be infinity, that we apply a minus sign by convention?

I just want to be sure exactly why the minus sign is introduced.

 

[swooshfactory]

Assuming your total charge is +Q, at the r=R your potential is negative (-Q/R). We know outside the charge distribution the potential should decrease in magnitude like 1/r. Thus infinity is not your zero of potential.

By analogy with the infinity is the zero of potential case, we know the potential should fall Q/R. So V(infinity) = -2Q/R.

Whatever you did is strange since your potential at the center of your sphere is negative. If I'm surrounded by a bunch of positive charge, why should the potential at my location be negative?

 

[Shinobii]

Well this is the equation I get for inside the sphere,

$$
3q \bigg( \int_{\infty}^R \frac{1}{r^2} dr + \int_R^r r \,\, dr \bigg)
$$

Which leaves me with the result stated. . .

 

[vanhees71]

I don't understand your equation. You have to evaluate the integral as given in your first posting:
\phi(\vec{x})=\int_{\mathbb{R}^3} \frac{\rho(\vec{x}')}{|\vec{x}'-\vec{x}|}.
For your example of a homgeneously charged sphere you have
\rho(\vec{x}')=\frac{3 Q}{4 \pi R^3} \Theta(R-|\vec{x}'|).
The integral is most easily performed in spherical coordinates with the polar axis in direction of \vec{x}, which gives
(\vec{x}'-\vec{x})^2=r'^2+r^2-2r r' \cos \vartheta'.
Here r=|\vec{x}| and =r&#039;=|\vec{x}&#039;| and \vartheta&amp;#039; is the polar angle. Then you have<br /> \phi(\vec{x})=\frac{3Q}{4 \pi R^3} \int_0^{2 \pi} \mathrm{d} \varphi&amp;#039; \int_0^{\pi} \mathrm{d} \vartheta&amp;#039; \int_0^R \mathrm{d} r&amp;#039; \; \frac{r&amp;#039;^2 \sin \vartheta&amp;#039;}{\sqrt{r^2+r&amp;#039;^2-2r r&amp;#039; \cos \vartheta&amp;#039;}}.<br /> Now you can do the angular integrations first. The one over \varphi&amp;#039; is trivial, just giving a factor 2 \pi. The one over \vartheta&amp;#039; is done by substitution u=\sin \vartheta&amp;#039;:<br /> \phi(\vec{x})=\frac{3Q}{2 R^3} \int_0^R \mathrm{d} r&amp;#039; \int_{-1}^1 \mathrm{d} u \frac{r&amp;#039;^2}{\sqrt{r^2+r&amp;#039;^2-2 r r&amp;#039; u}}=\frac{3Q}{2 R^3} \int_0^R \mathrm{d}r&amp;#039; \frac{r&amp;#039;}{r} (r+r&amp;#039;-|r-r&amp;#039;|).<br /> The remaining integral must be evaluated separately for the cases r&amp;lt;R and r&amp;gt;R.<br /> <br /> For r&amp;lt;R you have to split the integral in the region 0 \leq r&amp;#039; \leq r and r \leq r&amp;#039; \leq R, carefully evaluate the modulus and put the results together. The result is<br /> \phi(\vec{x})=\frac{Q}{2R^3} (3 R^2-r^2) \quad \text{for} \quad 0 \leq r \leq R.<br /> For r&amp;gt;R the integral is easy, because then always r&amp;#039;&amp;lt;r and thus |r-r&amp;#039;|=r-r&amp;#039;. This leads to<br /> \phi(\vec{x})=\frac{Q}{r} \quad \text{for} \quad r \geq R.<br /> As you see, the potential is continuous at r=R as it should be. <br /> <br /> Outside of the charge distribution, you get Coulomb&#039;s Law as if the whole charge is concentrated at the origin. This is true for any spherically symmetric charge distribution with compact support.

 

[Shinobii]

Ah I see now, I was going about it the wrong way it seems! Thank you very much @vanhees71 for the detailed solution, this clears up a lot of questions I had.

Also, (just in case you want to edit your post) you should have u = \cos(\theta&#039;). And you missed a / on your itex command (or an itex command altogether).