Edits to “Statement”, “Solution”
en/8.2.11.md
+2 −1
| @@ -1,8 +1,9 @@ | |||
| ### Statement | |||
| $8.2.11.$ An excess charge $Q_0$ is placed at the center of a conducting ball with resistivity $\rho$. | |||
| $\textbf{a. }$ How does the current flowing from the center of the ball to its surface depend on $Q_0$? | |||
| + | |||
| $\textbf{b. }$ How will the charge at the center of the ball change over time? | |||
| ### Solution | |||
| $\textbf{a. }$ Inside our conducting ball of radius $R$, let us draw a sphere of radius $r$ ($r < R$). The area of this sphere is $S = 4\pi r^2$. Let us write Gauss's law: | |||
| $$E(r) \cdot 4\pi r^2 = \frac{Q_0}{\varepsilon_0} \implies E(r) = \frac{Q_0}{4\pi\varepsilon_0 r^2}$$ | |||
| Ohm's law in differential form: | |||
| $$j(r) = \frac{1}{\rho} E(r)$$ | |||
| $$j(r) = \frac{1}{\rho} \cdot \frac{Q_0}{4\pi\varepsilon_0 r^2}$$ | |||
| where $j(r)$ is the current density. | |||
| The total current through the sphere of radius $r$: | |||
| @@ -22,7 +23,7 @@Solution | |||
| $$I = j(r) \cdot 4\pi r^2 = \frac{1}{\rho} \cdot \frac{Q_0}{4\pi\varepsilon_0 r^2} \cdot 4\pi r^2 = \frac{Q_0}{\varepsilon_0\rho}$$ | |||
| − | $\textbf{b. }$ Inside a conductor, the charge cannot remain concentrated: it will begin to flow to the surface, causing a current directed from the center to the surface. Using the result of subproblem $\textbf{a | ||
| + | $\textbf{b. }$ Inside a conductor, the charge cannot remain concentrated: it will begin to flow to the surface, causing a current directed from the center to the surface. Using the result of subproblem $\textbf{a }$, we can write: | ||
| $$I(t) = \frac{Q(t)}{\varepsilon_0\rho}$$ | |||
| But the current is the rate of decrease of charge: | |||
| $$I(t) = -\frac{dQ}{dt}$$ | |||
| Let us substitute: | |||
| $$-\frac{dQ}{dt} = \frac{Q}{\varepsilon_0\rho} \implies \frac{dQ}{Q} = -\frac{dt}{\varepsilon_0\rho}$$ | |||
| Integrating and then exponentiating: | |||
| $$Q(t) = C \cdot \exp\left(-\frac{t}{\varepsilon_0\rho}\right)$$ | |||
| From the initial condition $Q(0) = Q_0$, we obtain that $C = Q_0$. Then: | |||
| $$Q = Q_0 \exp\left(-\frac{t}{\varepsilon_0\rho}\right)$$ | |||
| #### Answer | |||
| a. $I = \frac{Q_0}{\varepsilon_0\rho}$ | |||
| b. $Q = Q_0 \exp\left(-\frac{t}{\varepsilon_0\rho}\right)$ | |||
| unchanged lines 21 | |||
| @@ -1,8 +1,9 @@ | |||
| ### Statement | ### Statement | ||
| $8.2.11.$ An excess charge $Q_0$ is placed at the center of a conducting ball with resistivity $\rho$. | $8.2.11.$ An excess charge $Q_0$ is placed at the center of a conducting ball with resistivity $\rho$. | ||
| $\textbf{a. }$ How does the current flowing from the center of the ball to its surface depend on $Q_0$? | $\textbf{a. }$ How does the current flowing from the center of the ball to its surface depend on $Q_0$? | ||
| $\textbf{b. }$ How will the charge at the center of the ball change over time? | $\textbf{b. }$ How will the charge at the center of the ball change over time? | ||
| ### Solution | ### Solution | ||
| $\textbf{a. }$ Inside our conducting ball of radius $R$, let us draw a sphere of radius $r$ ($r < R$). The area of this sphere is $S = 4\pi r^2$. Let us write Gauss's law: | $\textbf{a. }$ Inside our conducting ball of radius $R$, let us draw a sphere of radius $r$ ($r < R$). The area of this sphere is $S = 4\pi r^2$. Let us write Gauss's law: | ||
| $$E(r) \cdot 4\pi r^2 = \frac{Q_0}{\varepsilon_0} \implies E(r) = \frac{Q_0}{4\pi\varepsilon_0 r^2}$$ | $$E(r) \cdot 4\pi r^2 = \frac{Q_0}{\varepsilon_0} \implies E(r) = \frac{Q_0}{4\pi\varepsilon_0 r^2}$$ | ||
| Ohm's law in differential form: | Ohm's law in differential form: | ||
| $$j(r) = \frac{1}{\rho} E(r)$$ | $$j(r) = \frac{1}{\rho} E(r)$$ | ||
| $$j(r) = \frac{1}{\rho} \cdot \frac{Q_0}{4\pi\varepsilon_0 r^2}$$ | $$j(r) = \frac{1}{\rho} \cdot \frac{Q_0}{4\pi\varepsilon_0 r^2}$$ | ||
| where $j(r)$ is the current density. | where $j(r)$ is the current density. | ||
| The total current through the sphere of radius $r$: | The total current through the sphere of radius $r$: | ||
| @@ -22,7 +23,7 @@Solution | |||
| $$I = j(r) \cdot 4\pi r^2 = \frac{1}{\rho} \cdot \frac{Q_0}{4\pi\varepsilon_0 r^2} \cdot 4\pi r^2 = \frac{Q_0}{\varepsilon_0\rho}$$ | $$I = j(r) \cdot 4\pi r^2 = \frac{1}{\rho} \cdot \frac{Q_0}{4\pi\varepsilon_0 r^2} \cdot 4\pi r^2 = \frac{Q_0}{\varepsilon_0\rho}$$ | ||
| $\textbf{b. }$ Inside a conductor, the charge cannot remain concentrated: it will begin to flow to the surface, causing a current directed from the center to the surface. Using the result of subproblem $\textbf{a |
$\textbf{b. }$ Inside a conductor, the charge cannot remain concentrated: it will begin to flow to the surface, causing a current directed from the center to the surface. Using the result of subproblem $\textbf{a }$, we can write: | ||
| $$I(t) = \frac{Q(t)}{\varepsilon_0\rho}$$ | $$I(t) = \frac{Q(t)}{\varepsilon_0\rho}$$ | ||
| But the current is the rate of decrease of charge: | But the current is the rate of decrease of charge: | ||
| $$I(t) = -\frac{dQ}{dt}$$ | $$I(t) = -\frac{dQ}{dt}$$ | ||
| Let us substitute: | Let us substitute: | ||
| $$-\frac{dQ}{dt} = \frac{Q}{\varepsilon_0\rho} \implies \frac{dQ}{Q} = -\frac{dt}{\varepsilon_0\rho}$$ | $$-\frac{dQ}{dt} = \frac{Q}{\varepsilon_0\rho} \implies \frac{dQ}{Q} = -\frac{dt}{\varepsilon_0\rho}$$ | ||
| Integrating and then exponentiating: | Integrating and then exponentiating: | ||
| $$Q(t) = C \cdot \exp\left(-\frac{t}{\varepsilon_0\rho}\right)$$ | $$Q(t) = C \cdot \exp\left(-\frac{t}{\varepsilon_0\rho}\right)$$ | ||
| From the initial condition $Q(0) = Q_0$, we obtain that $C = Q_0$. Then: | From the initial condition $Q(0) = Q_0$, we obtain that $C = Q_0$. Then: | ||
| $$Q = Q_0 \exp\left(-\frac{t}{\varepsilon_0\rho}\right)$$ | $$Q = Q_0 \exp\left(-\frac{t}{\varepsilon_0\rho}\right)$$ | ||
| #### Answer | #### Answer | ||
| a. $I = \frac{Q_0}{\varepsilon_0\rho}$ | a. $I = \frac{Q_0}{\varepsilon_0\rho}$ | ||
| b. $Q = Q_0 \exp\left(-\frac{t}{\varepsilon_0\rho}\right)$ | b. $Q = Q_0 \exp\left(-\frac{t}{\varepsilon_0\rho}\right)$ | ||
| unchanged lines 21 | |||