Edit to “Solution”

Luisito edited Jun 9, 2026, 9:53 PM

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			### Statement

			$6.3.7.$ A sphere of radius $R$ has charge $Q$. What is the potential of the field in the center of the sphere? Does the potential at the center of the sphere depend on the distribution of charges on the sphere? Does the field potential on the surface of a sphere depend on the charge distribution over the sphere?

		@@ -5,7 +5,7 @@Statement
			### Solution

			As the statement of the problem doesn't clarify if the sphere is conducting or non-conductive, let's consider both cases.\
		−	Case 1) Nin-Conductive sphere:\
		+	Case 1) Non-Conductive sphere:\
			Let's assume that charge is uniformly distributed in the volume. Let electrical potential be:\
			$V(r) = -\int_{\infty}^{r} \vec{E} \cdot d\vec{r}$ (1)\
			for the center, let's consider $r<<R$, or $r \rightarrow 0$.\
			Applying Gauss Law:\
			For $r<R$,\
			$\int \vec{E} \cdot d\vec{S} = \frac{q_{enc}}{\varepsilon_0}$\
			but the enclosed charge into a sphere of radius r is related to the charge distribution per unit of volume,\
			$\rho = \frac{Q}{\frac{4}{3}\pi R^3} = \frac{q_{enc}}{\frac{4}{3}\pi r^3}$\
			$q_{enc} = Q\left(\frac{r}{R}\right)^3$\
			so,\
			$E(r) = \frac{Q r}{4\pi\varepsilon_0 R^3}\;\;\;\;\forall\;r < R$\
			and for $r > R$, the enclosed charge is Q, then,\
			$E(r) = \frac{Q}{4\pi\varepsilon_0 r^2}\;\;\;\;\forall\;r>R$ (2)\
			This mean that function $E(r)$ has two behaviors, depending on values of $r$. According (1) and assuming $r\rightarrow 0$,\
			$V(r) = -\left(\int_{\infty}^{R} \vec{E} \cdot d\vec{r} + \int_{R}^{r} \vec{E} \cdot d\vec{r}\right)$ (I)\
			developing,\
			$V(r) = \frac{Q}{4\pi\varepsilon_0 R} - \frac{Q}{8\pi\varepsilon_0 R^3}(r^2-R^2)$\
			as $r$ tends to zero,\
			$V(r) = \frac{3Q}{8\pi\varepsilon_0 R^3}$ (3)\
			Case 2) Conducting sphere\
			In this case, charge stays on sphere surface, so the electric field inside of it is null ($E(r) = 0$), so potential is constant inside the sphere and coincides with the value of it on the surface. According to (1) and (2),\
			$V(r) = -\int_{\infty}^{R} \frac{Q}{4\pi\varepsilon_0 r^2}dr$\
			$V(0) = V(r) = \frac{Q}{4\pi\varepsilon_0 R}$ (4)\
			Finally, if sphere is non-conductive $V(0)$ depends on the charge distribution, as we saw in (I), but in the second case (maybe the problem refers specifically to a conducting sphere), it doesn't depend on it. If the charge is non-uniformly distributed, the potential on the surface does change with the local distribution. For example, charge accumulations in certain areas generate angular variations in the potential. However, outside the sphere, at a great distance, the potential depends only on the total charge, Q, as if it were a point charge at the center.

			#### Answer

			$V(0) = \frac{Q}{4\pi\varepsilon_0 R}$, NO, YES.
		unchanged lines 25

	### Statement		### Statement

	$6.3.7.$ A sphere of radius $R$ has charge $Q$. What is the potential of the field in the center of the sphere? Does the potential at the center of the sphere depend on the distribution of charges on the sphere? Does the field potential on the surface of a sphere depend on the charge distribution over the sphere?		$6.3.7.$ A sphere of radius $R$ has charge $Q$. What is the potential of the field in the center of the sphere? Does the potential at the center of the sphere depend on the distribution of charges on the sphere? Does the field potential on the surface of a sphere depend on the charge distribution over the sphere?

	@@ -5,7 +5,7 @@Statement
	### Solution		### Solution

	As the statement of the problem doesn't clarify if the sphere is conducting or non-conductive, let's consider both cases.\		As the statement of the problem doesn't clarify if the sphere is conducting or non-conductive, let's consider both cases.\
	Case 1) Nin-Conductive sphere:\		Case 1) Non-Conductive sphere:\
	Let's assume that charge is uniformly distributed in the volume. Let electrical potential be:\		Let's assume that charge is uniformly distributed in the volume. Let electrical potential be:\
	$V(r) = -\int_{\infty}^{r} \vec{E} \cdot d\vec{r}$ (1)\		$V(r) = -\int_{\infty}^{r} \vec{E} \cdot d\vec{r}$ (1)\
	for the center, let's consider $r<<R$, or $r \rightarrow 0$.\		for the center, let's consider $r<<R$, or $r \rightarrow 0$.\
	Applying Gauss Law:\		Applying Gauss Law:\
	For $r<R$,\		For $r<R$,\
	$\int \vec{E} \cdot d\vec{S} = \frac{q_{enc}}{\varepsilon_0}$\		$\int \vec{E} \cdot d\vec{S} = \frac{q_{enc}}{\varepsilon_0}$\
	but the enclosed charge into a sphere of radius r is related to the charge distribution per unit of volume,\		but the enclosed charge into a sphere of radius r is related to the charge distribution per unit of volume,\
	$\rho = \frac{Q}{\frac{4}{3}\pi R^3} = \frac{q_{enc}}{\frac{4}{3}\pi r^3}$\		$\rho = \frac{Q}{\frac{4}{3}\pi R^3} = \frac{q_{enc}}{\frac{4}{3}\pi r^3}$\
	$q_{enc} = Q\left(\frac{r}{R}\right)^3$\		$q_{enc} = Q\left(\frac{r}{R}\right)^3$\
	so,\		so,\
	$E(r) = \frac{Q r}{4\pi\varepsilon_0 R^3}\;\;\;\;\forall\;r < R$\		$E(r) = \frac{Q r}{4\pi\varepsilon_0 R^3}\;\;\;\;\forall\;r < R$\
	and for $r > R$, the enclosed charge is Q, then,\		and for $r > R$, the enclosed charge is Q, then,\
	$E(r) = \frac{Q}{4\pi\varepsilon_0 r^2}\;\;\;\;\forall\;r>R$ (2)\		$E(r) = \frac{Q}{4\pi\varepsilon_0 r^2}\;\;\;\;\forall\;r>R$ (2)\
	This mean that function $E(r)$ has two behaviors, depending on values of $r$. According (1) and assuming $r\rightarrow 0$,\		This mean that function $E(r)$ has two behaviors, depending on values of $r$. According (1) and assuming $r\rightarrow 0$,\
	$V(r) = -\left(\int_{\infty}^{R} \vec{E} \cdot d\vec{r} + \int_{R}^{r} \vec{E} \cdot d\vec{r}\right)$ (I)\		$V(r) = -\left(\int_{\infty}^{R} \vec{E} \cdot d\vec{r} + \int_{R}^{r} \vec{E} \cdot d\vec{r}\right)$ (I)\
	developing,\		developing,\
	$V(r) = \frac{Q}{4\pi\varepsilon_0 R} - \frac{Q}{8\pi\varepsilon_0 R^3}(r^2-R^2)$\		$V(r) = \frac{Q}{4\pi\varepsilon_0 R} - \frac{Q}{8\pi\varepsilon_0 R^3}(r^2-R^2)$\
	as $r$ tends to zero,\		as $r$ tends to zero,\
	$V(r) = \frac{3Q}{8\pi\varepsilon_0 R^3}$ (3)\		$V(r) = \frac{3Q}{8\pi\varepsilon_0 R^3}$ (3)\
	Case 2) Conducting sphere\		Case 2) Conducting sphere\
	In this case, charge stays on sphere surface, so the electric field inside of it is null ($E(r) = 0$), so potential is constant inside the sphere and coincides with the value of it on the surface. According to (1) and (2),\		In this case, charge stays on sphere surface, so the electric field inside of it is null ($E(r) = 0$), so potential is constant inside the sphere and coincides with the value of it on the surface. According to (1) and (2),\
	$V(r) = -\int_{\infty}^{R} \frac{Q}{4\pi\varepsilon_0 r^2}dr$\		$V(r) = -\int_{\infty}^{R} \frac{Q}{4\pi\varepsilon_0 r^2}dr$\
	$V(0) = V(r) = \frac{Q}{4\pi\varepsilon_0 R}$ (4)\		$V(0) = V(r) = \frac{Q}{4\pi\varepsilon_0 R}$ (4)\
	Finally, if sphere is non-conductive $V(0)$ depends on the charge distribution, as we saw in (I), but in the second case (maybe the problem refers specifically to a conducting sphere), it doesn't depend on it. If the charge is non-uniformly distributed, the potential on the surface does change with the local distribution. For example, charge accumulations in certain areas generate angular variations in the potential. However, outside the sphere, at a great distance, the potential depends only on the total charge, Q, as if it were a point charge at the center.		Finally, if sphere is non-conductive $V(0)$ depends on the charge distribution, as we saw in (I), but in the second case (maybe the problem refers specifically to a conducting sphere), it doesn't depend on it. If the charge is non-uniformly distributed, the potential on the surface does change with the local distribution. For example, charge accumulations in certain areas generate angular variations in the potential. However, outside the sphere, at a great distance, the potential depends only on the total charge, Q, as if it were a point charge at the center.

	#### Answer		#### Answer

	$V(0) = \frac{Q}{4\pi\varepsilon_0 R}$, NO, YES.		$V(0) = \frac{Q}{4\pi\varepsilon_0 R}$, NO, YES.
	unchanged lines 25