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Tete edited Aug 15, 2026, 8:25 AM

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			### Statement

			$6.3.32.$ What is the charge induced on the surface of a grounded metal ball by a point charge $q$ located at a distance $L$ from the center of the ball? Ball radius $R<L$.

			### Solution
		@@ -6,7 +6,7 @@Solution

			![For problem $6.3.32$ \|790x440, 100%](../../img/6.3.32/Savchenko.png)

		−	From Gauss' Theorem, the question is equivalent to asking ~~where, w~~ithout the metal ball~~, a~~ point charge $-Q$ can be placed so that together with the point charge $+q$ a zero-potential surface is created where the metal ball used to be. By symmetry, the point charge $-Q$ must be placed on a line joining the point charge $+q$ and where the center of the metal ball used to be. As a result, we may solve this problem in two dimensions, i.e. considering a zero-potential circle instead of a zero-potential sphere. Let the point charge $+q$ be at point $Y$ on the extension of the diameter $AB$ of the circle and let $C$ be any point on the circle. We claim that the point charge $-Q$ must be placed at point $X$ on $AB$ so that $\angle XCB=\angle YCB$. Since $\angle ACB=90^\circ$, it follows that $\angle XCA=\angle DCA$ as well. By angle-bisector theorems,
		+	From Gauss' Theorem, the question is equivalent to asking (without the metal ball) what point charge $-Q$ can be placed somewhere so that together with the point charge $+q$ a zero-potential surface is created where the metal ball used to be. By symmetry, the point charge $-Q$ must be placed on a line joining the point charge $+q$ and where the center of the metal ball used to be. As a result, we may solve this problem in two dimensions, i.e. considering a zero-potential circle instead of a zero-potential sphere. Let the point charge $+q$ be at point $Y$ on the extension of the diameter $AB$ of the circle and let $C$ be any point on the circle. We claim that the point charge $-Q$ must be placed at point $X$ on $AB$ so that $\angle XCB=\angle YCB$. Since $\angle ACB=90^\circ$, it follows that $\angle XCA=\angle DCA$ as well. By angle-bisector theorems,

			\[\frac{XB}{YB}=\frac{XC}{YC}=\frac{XA}{YA}\qquad\Rightarrow\qquad\frac{XB}{L-R}=\frac{XA}{L+R}.\]

			Together with the fact that $XA+XB=2R$, we get $XA=R(L+R)/L$, which is independent of point $C$. As $XC/YC=XA/YA=R/L$ and electric potential from a point charge varies inversely with the distance from the point charge, we choose $-Q=-qR/L$ to guarantee that the electric potential due to point charges $+q$ and $-Q$ are identically zero on the circle.

			#### Answer

			$-qR/L$
		unchanged lines 5

	### Statement		### Statement

	$6.3.32.$ What is the charge induced on the surface of a grounded metal ball by a point charge $q$ located at a distance $L$ from the center of the ball? Ball radius $R<L$.		$6.3.32.$ What is the charge induced on the surface of a grounded metal ball by a point charge $q$ located at a distance $L$ from the center of the ball? Ball radius $R<L$.

	### Solution		### Solution
	@@ -6,7 +6,7 @@Solution

	![For problem $6.3.32$ \|790x440, 100%](../../img/6.3.32/Savchenko.png)		![For problem $6.3.32$ \|790x440, 100%](../../img/6.3.32/Savchenko.png)

	From Gauss' Theorem, the question is equivalent to asking ~~where, w~~ithout the metal ball~~, a~~ point charge $-Q$ can be placed so that together with the point charge $+q$ a zero-potential surface is created where the metal ball used to be. By symmetry, the point charge $-Q$ must be placed on a line joining the point charge $+q$ and where the center of the metal ball used to be. As a result, we may solve this problem in two dimensions, i.e. considering a zero-potential circle instead of a zero-potential sphere. Let the point charge $+q$ be at point $Y$ on the extension of the diameter $AB$ of the circle and let $C$ be any point on the circle. We claim that the point charge $-Q$ must be placed at point $X$ on $AB$ so that $\angle XCB=\angle YCB$. Since $\angle ACB=90^\circ$, it follows that $\angle XCA=\angle DCA$ as well. By angle-bisector theorems,		From Gauss' Theorem, the question is equivalent to asking (without the metal ball) what point charge $-Q$ can be placed somewhere so that together with the point charge $+q$ a zero-potential surface is created where the metal ball used to be. By symmetry, the point charge $-Q$ must be placed on a line joining the point charge $+q$ and where the center of the metal ball used to be. As a result, we may solve this problem in two dimensions, i.e. considering a zero-potential circle instead of a zero-potential sphere. Let the point charge $+q$ be at point $Y$ on the extension of the diameter $AB$ of the circle and let $C$ be any point on the circle. We claim that the point charge $-Q$ must be placed at point $X$ on $AB$ so that $\angle XCB=\angle YCB$. Since $\angle ACB=90^\circ$, it follows that $\angle XCA=\angle DCA$ as well. By angle-bisector theorems,

	\[\frac{XB}{YB}=\frac{XC}{YC}=\frac{XA}{YA}\qquad\Rightarrow\qquad\frac{XB}{L-R}=\frac{XA}{L+R}.\]		\[\frac{XB}{YB}=\frac{XC}{YC}=\frac{XA}{YA}\qquad\Rightarrow\qquad\frac{XB}{L-R}=\frac{XA}{L+R}.\]

	Together with the fact that $XA+XB=2R$, we get $XA=R(L+R)/L$, which is independent of point $C$. As $XC/YC=XA/YA=R/L$ and electric potential from a point charge varies inversely with the distance from the point charge, we choose $-Q=-qR/L$ to guarantee that the electric potential due to point charges $+q$ and $-Q$ are identically zero on the circle.		Together with the fact that $XA+XB=2R$, we get $XA=R(L+R)/L$, which is independent of point $C$. As $XC/YC=XA/YA=R/L$ and electric potential from a point charge varies inversely with the distance from the point charge, we choose $-Q=-qR/L$ to guarantee that the electric potential due to point charges $+q$ and $-Q$ are identically zero on the circle.

	#### Answer		#### Answer

	$-qR/L$		$-qR/L$
	unchanged lines 5