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Tete edited Sep 12, 2026, 4:33 PM

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		+	### Statement
		+
		+	$6.5.18.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	In electrostatics, the energy stored in the electric field produced by a charge distribution is equal to the potential energy of the distribution. If a ball of radius $r$ has uniform charge density $\rho$, then the potential at its surface is
		+
		+	\[V=\frac{k}{r}\cdot\rho\cdot\frac{4}{3}\pi r^3=\frac{\rho r^2}{3\epsilon_0}.\]
		+
		+	If a thin spherical shell of charge $\rho\cdot4\pi r^2\,dr$ is added to its surface, then the potential energy of the system increases by
		+
		+	\[dU=V\cdot\rho\cdot4\pi r^2\,dr=\frac{4\pi\rho^2r^4}{3\epsilon_0}.\]
		+
		+	(The potential energy in the thin spherical shell is negligible compared to $dU$.) Thus,
		+
		+	\[U=\frac{4\pi\rho^2r^5}{15\epsilon_0}=\frac{3k}{5r}\left(\rho\cdot\frac{4}{3}\pi r^3\right)^2.\]
		+
		+	Using $R$ for $r$ and $Q$ for $\rho\cdot4\pi R^3/3$, we have $U=3kQ^2/(5r)$.
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

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			### Statement

			$6.5.18.$ [Insert the problem statement]

			### Solution

			In electrostatics, the energy stored in the electric field produced by a charge distribution is equal to the potential energy of the distribution. If a ball of radius $r$ has uniform charge density $\rho$, then the potential at its surface is

			\[V=\frac{k}{r}\cdot\rho\cdot\frac{4}{3}\pi r^3=\frac{\rho r^2}{3\epsilon_0}.\]

			If a thin spherical shell of charge $\rho\cdot4\pi r^2\,dr$ is added to its surface, then the potential energy of the system increases by

			\[dU=V\cdot\rho\cdot4\pi r^2\,dr=\frac{4\pi\rho^2r^4}{3\epsilon_0}.\]

			(The potential energy in the thin spherical shell is negligible compared to $dU$.) Thus,

			\[U=\frac{4\pi\rho^2r^5}{15\epsilon_0}=\frac{3k}{5r}\left(\rho\cdot\frac{4}{3}\pi r^3\right)^2.\]

			Using $R$ for $r$ and $Q$ for $\rho\cdot4\pi R^3/3$, we have $U=3kQ^2/(5r)$.

			#### Answer

			[Insert a concise answer or boxed result]