Edits to “Statement”, “Solution”, “Answer”
en/6.5.18.md
+4 −4
| @@ -1,6 +1,6 @@ | |||
| ### Statement | |||
| − | $6.5.18.$ | ||
| + | $6.5.18.$ Determine the energy of the electric field of a uniformly charged ball of radius $R$. Full charge of ball $Q$. | ||
| ### 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 | |||
| \[V=\frac{k}{r}\cdot\rho\cdot\frac{4}{3}\pi r^3=\frac{\rho r^2}{3\epsilon_0}.\] | |||
| @@ -10,14 +10,14 @@Solution | |||
| 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 | |||
| − | \[dU=V\cdot\rho\cdot4\pi r^2\,dr=\frac{4\pi\rho^2r^4}{3\epsilon_0}.\] | ||
| + | \[dU=V\cdot\rho\cdot4\pi r^2\,dr=\frac{4\pi\rho^2r^4}{3\epsilon_0}\,dr.\] | ||
| (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/(5 | ||
| + | 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] | ||
| + | $\frac{3kQ^2}{5R}$ | ||
| @@ -1,6 +1,6 @@ | |||
| ### Statement | ### Statement | ||
| $6.5.18.$ |
$6.5.18.$ Determine the energy of the electric field of a uniformly charged ball of radius $R$. Full charge of ball $Q$. | ||
| ### Solution | ### 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 | 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 | ||
| \[V=\frac{k}{r}\cdot\rho\cdot\frac{4}{3}\pi r^3=\frac{\rho r^2}{3\epsilon_0}.\] | \[V=\frac{k}{r}\cdot\rho\cdot\frac{4}{3}\pi r^3=\frac{\rho r^2}{3\epsilon_0}.\] | ||
| @@ -10,14 +10,14 @@Solution | |||
| 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 | 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 | ||
| \[dU=V\cdot\rho\cdot4\pi r^2\,dr=\frac{4\pi\rho^2r^4}{3\epsilon_0}.\] | \[dU=V\cdot\rho\cdot4\pi r^2\,dr=\frac{4\pi\rho^2r^4}{3\epsilon_0}\,dr.\] | ||
| (The potential energy in the thin spherical shell is negligible compared to $dU$.) Thus, | (The potential energy in the thin spherical shell is negligible compared to $dU$.) Thus, | ||
| \[U=\frac{4\pi\rho^2r^5}{15\epsilon_0}=\frac{3k}{5r}\left(\rho\cdot\frac{4}{3}\pi r^3\right)^2.\] | \[U=\frac{4\pi\rho^2r^5}{15\epsilon_0}=\frac{3k}{5r}\left(\rho\cdot\frac{4}{3}\pi r^3\right)^2.\] | ||
| Using $R$ for $r$ and $Q$ for $\rho\cdot4\pi R^3/3$, we have $U=3kQ^2/(5 |
Using $R$ for $r$ and $Q$ for $\rho\cdot4\pi R^3/3$, we have $U=3kQ^2/(5R)$. | ||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $\frac{3kQ^2}{5R}$ | ||