Правка разделов «Solution», «Answer»
en/6.5.9.md
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| ### Statement | |||
| $6.5.9^*.$ A uniformly charged sphere of radius $R$ is cut into two parts along a plane spaced at a distance $h$ from the center of the sphere. Find the force with which these parts repel each other. Full charge of the sphere $Q$. What is the minimum charge you need to put in the center of the sphere so that its parts do not fly apart? | |||
| ### Solution | |||
| @@ -6,12 +6,14 @@Solution | |||
|  | |||
| − | The electric field at the surface of a uniformly charged sphere is $Q/(4\pi\epsilon_0R^2)=\sigma/\epsilon_0$, where $\sigma$ is the surface charge density. Since a small piece of the sphere contributes $\sigma/(2\epsilon_0)$ of local electric field, this means that the field due to the sphere minus that piece is also $\sigma/(2\epsilon_0)$. If that small piece (shown in red) has area $a$, then it has charge $\sigma a$ and experience an outward force of magnitude $\sigma^2a/(2\epsilon_0)$. Due to the symmetry of the problem, only the vertical component $\sigma^2a\cos\alpha/(2\epsilon_0)$ of this force contributes to the total force to be determined. Since $a\cos\alpha$ is the area of the horizontal projection (shown in green) of this small piece, the total force on the upper part of the sphere is the product of $\sigma^2/(2\epsilon_0)$ and the area of its horizontal projection $\pi(R^2-h^2)$. We have | ||
| + | The electric field at the surface of a uniformly charged sphere is $Q/(4\pi\epsilon_0R^2)=\sigma/\epsilon_0$, where $\sigma=Q/(4\pi R^2)$ is the surface charge density. Since a small piece of the sphere contributes $\sigma/(2\epsilon_0)$ of local electric field, this means that the field due to the sphere minus that piece is also $\sigma/(2\epsilon_0)$. If that small piece (shown in red) has area $a$, then it has charge $\sigma a$ and experiences an outward force of magnitude $\sigma^2a/(2\epsilon_0)$. Due to the symmetry of the problem, only the vertical component $\sigma^2a\cos\alpha/(2\epsilon_0)$ of this force contributes to the total force to be determined. Since $a\cos\alpha$ is the area of the horizontal projection (shown in green) of this small piece, the total force on the upper part of the sphere is the product of $\sigma^2/(2\epsilon_0)$ and the area of its horizontal projection $\pi(R^2-h^2)$. We have | ||
| \[F=\frac{\sigma^2}{2\epsilon_0}\cdot\pi(R^2-h^2)=\frac{Q^2(R^2-h^2)}{32\pi\epsilon_0R^4}.\] | |||
| − | To hold the two parts of the sphere together, we need to put a | ||
| + | To hold the two parts of the sphere together, we need to put a negative charge no smaller than $-Q/2$ at the center of the sphere to produce the electric field that overcome the field $\sigma/(2\epsilon_0)$ on each small piece of the sphere. | ||
| #### Answer | |||
| $F=\frac{Q^2(R^2-h^2)}{32\pi\epsilon_0R^4}$ | |||
| + | |||
| + | A negative charge no smaller than $-Q/2$. | ||
| ### Statement | ### Statement | ||
| $6.5.9^*.$ A uniformly charged sphere of radius $R$ is cut into two parts along a plane spaced at a distance $h$ from the center of the sphere. Find the force with which these parts repel each other. Full charge of the sphere $Q$. What is the minimum charge you need to put in the center of the sphere so that its parts do not fly apart? | $6.5.9^*.$ A uniformly charged sphere of radius $R$ is cut into two parts along a plane spaced at a distance $h$ from the center of the sphere. Find the force with which these parts repel each other. Full charge of the sphere $Q$. What is the minimum charge you need to put in the center of the sphere so that its parts do not fly apart? | ||
| ### Solution | ### Solution | ||
| @@ -6,12 +6,14 @@Solution | |||
|  |  | ||
| The electric field at the surface of a uniformly charged sphere is $Q/(4\pi\epsilon_0R^2)=\sigma/\epsilon_0$, where $\sigma$ is the surface charge density. Since a small piece of the sphere contributes $\sigma/(2\epsilon_0)$ of local electric field, this means that the field due to the sphere minus that piece is also $\sigma/(2\epsilon_0)$. If that small piece (shown in red) has area $a$, then it has charge $\sigma a$ and experience an outward force of magnitude $\sigma^2a/(2\epsilon_0)$. Due to the symmetry of the problem, only the vertical component $\sigma^2a\cos\alpha/(2\epsilon_0)$ of this force contributes to the total force to be determined. Since $a\cos\alpha$ is the area of the horizontal projection (shown in green) of this small piece, the total force on the upper part of the sphere is the product of $\sigma^2/(2\epsilon_0)$ and the area of its horizontal projection $\pi(R^2-h^2)$. We have | The electric field at the surface of a uniformly charged sphere is $Q/(4\pi\epsilon_0R^2)=\sigma/\epsilon_0$, where $\sigma=Q/(4\pi R^2)$ is the surface charge density. Since a small piece of the sphere contributes $\sigma/(2\epsilon_0)$ of local electric field, this means that the field due to the sphere minus that piece is also $\sigma/(2\epsilon_0)$. If that small piece (shown in red) has area $a$, then it has charge $\sigma a$ and experiences an outward force of magnitude $\sigma^2a/(2\epsilon_0)$. Due to the symmetry of the problem, only the vertical component $\sigma^2a\cos\alpha/(2\epsilon_0)$ of this force contributes to the total force to be determined. Since $a\cos\alpha$ is the area of the horizontal projection (shown in green) of this small piece, the total force on the upper part of the sphere is the product of $\sigma^2/(2\epsilon_0)$ and the area of its horizontal projection $\pi(R^2-h^2)$. We have | ||
| \[F=\frac{\sigma^2}{2\epsilon_0}\cdot\pi(R^2-h^2)=\frac{Q^2(R^2-h^2)}{32\pi\epsilon_0R^4}.\] | \[F=\frac{\sigma^2}{2\epsilon_0}\cdot\pi(R^2-h^2)=\frac{Q^2(R^2-h^2)}{32\pi\epsilon_0R^4}.\] | ||
| To hold the two parts of the sphere together, we need to put a |
To hold the two parts of the sphere together, we need to put a negative charge no smaller than $-Q/2$ at the center of the sphere to produce the electric field that overcome the field $\sigma/(2\epsilon_0)$ on each small piece of the sphere. | ||
| #### Answer | #### Answer | ||
| $F=\frac{Q^2(R^2-h^2)}{32\pi\epsilon_0R^4}$ | $F=\frac{Q^2(R^2-h^2)}{32\pi\epsilon_0R^4}$ | ||
| A negative charge no smaller than $-Q/2$. | |||