# Potential energy for uniform sphere doubt

For a sphere of radius $$R$$, with electric charge distributed in uniform way $$\rho=Q/V$$, we have

$$U_e=\dfrac{3K_CQ^2}{5R}$$

However, when substituting the charge $$Q=\rho V=\rho 4\pi R^3/3$$, we get

$$U_e=\dfrac{16\pi^2 K_C\rho^2 R^5}{15}$$

My question is simple, which one gives the right electrostatic value as $$R\rightarrow 0$$ and $$R\rightarrow \infty$$? In the former case, the first expression provides $$U_e=\infty$$ but the latter gives $$U_e=0$$! In the case of big $$R$$, is the total energy inversely proportional to $$R$$ or directly proportional to $$R^5$$? Or both since, as $$R\rightarrow \infty$$, then $$\rho\rightarrow 0$$, and $$R\rightarrow 0$$ implies $$\rho\rightarrow \infty$$. But if this is the case, which is the correct power of proportionality for energy? Is energy directly proportional to $$R^n$$ or inversely proportional to $$R$$?

• Don't forget that in the second equation, you also have a dependency of $\rho$ of $R$. Commented Apr 12, 2022 at 13:02
• Yep, but then, are both correct? I find disturbing one says the energy grows with the size, and the another that energy decreases with the size R. Which one is correct???? Commented Apr 12, 2022 at 15:19
• Both equations are correct. Taking $R\rightarrow 0$ or $R\rightarrow\infty$, you have to consider $\rho(R)$ in the second equation, and putting that dependency back in yields that of the first equation, so the limit for both equations is the same, that of the first one. Commented Apr 12, 2022 at 15:24

Since your $$Q(R)=\rho V(R)=\frac{\rho 4\pi R^3}{3}$$ is function of $$R$$ so$$\displaystyle\lim_{R\to\infty} U_e=\displaystyle\lim_{R\to\infty}\dfrac{3K_C(Q(R))^2}{5R}=\displaystyle\lim_{R\to\infty}\dfrac{16\pi^2 K_C\rho^2 R^5}{15}=\infty$$.