Новое решение
en/6.3.32.md
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| + | ### Statement | ||
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| + | $6.3.32.$ [Insert the problem statement] | ||
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| + | ### Solution | ||
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| + | From Gauss' Theorem, the question is equivalent to asking where, without 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, | ||
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| + | \[\frac{XB}{YB}=\frac{XC}{YC}=\frac{XA}{YA}\qquad\Rightarrow\qquad\frac{XB}{L-R}=\frac{XA}{L+R}.\] | ||
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| + | 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. | ||
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| + | #### Answer | ||
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| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $6.3.32.$ [Insert the problem statement] | |||
| ### Solution | |||
|  | |||
| From Gauss' Theorem, the question is equivalent to asking where, without 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, | |||
| \[\frac{XB}{YB}=\frac{XC}{YC}=\frac{XA}{YA}\qquad\Rightarrow\qquad\frac{XB}{L-R}=\frac{XA}{L+R}.\] | |||
| 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 | |||
| [Insert a concise answer or boxed result] | |||