Edit to “Solution”
en/8.2.24.md
+1 −1
| ### Statement | |||
| $8.2.24.$ The tape consists of narrow conductive strips with even narrower insulating gaps. It is in contact with one plate of the capacitor and with a small contact, between which the resistance $R$ is turned on. Before that, the tape was not charged, and the charge of the capacitor plates was $\pm q$. The length of the plates is $l$, and their width coincides with the width of the tape. The tape is pulled out of the condenser with a force of $F$. Find the current through the resistance and the steady-state speed of the tape. The distance from the edge of the capacitor plates to the contact is much greater than the distance between them and much less than their length. | |||
| ### Solution | |||
| @@ -6,7 +6,7 @@Solution | |||
|  | |||
| − | Suppose the steady-state speed of the tape is $v$. Then, in a short time interval $\Delta t$ the charge $-qv\Delta t/l$ is removed from the region between the plates upward along the tape. Equivalently, conventional current $I=qv/l$ flows upward through the resistor, which generates heat at the rate of $I^2R=q^2v^2R/l^2$. Since this must equal | ||
| + | Suppose the steady-state speed of the tape is $v$. Then, in a short time interval $\Delta t$ the charge $-qv\Delta t/l$ is removed from the region between the plates upward along the tape. Equivalently, conventional current $I=qv/l$ flows upward through the resistor, which generates heat at the rate of $I^2R=q^2v^2R/l^2$. Since this must equal the power input $Fv$ of the system, we have | ||
| \[\frac{q^2v^2R}{l^2}=Fv\qquad\Rightarrow\qquad v=\frac{Fl^2}{q^2R},\] | |||
| and thus $I=Fl/(qR)$. | |||
| #### Answer | |||
| $I=\frac{Fl}{qR}$, $v=\frac{Fl^2}{q^2R}$ | |||
| unchanged lines 5 | |||
| ### Statement | ### Statement | ||
| $8.2.24.$ The tape consists of narrow conductive strips with even narrower insulating gaps. It is in contact with one plate of the capacitor and with a small contact, between which the resistance $R$ is turned on. Before that, the tape was not charged, and the charge of the capacitor plates was $\pm q$. The length of the plates is $l$, and their width coincides with the width of the tape. The tape is pulled out of the condenser with a force of $F$. Find the current through the resistance and the steady-state speed of the tape. The distance from the edge of the capacitor plates to the contact is much greater than the distance between them and much less than their length. | $8.2.24.$ The tape consists of narrow conductive strips with even narrower insulating gaps. It is in contact with one plate of the capacitor and with a small contact, between which the resistance $R$ is turned on. Before that, the tape was not charged, and the charge of the capacitor plates was $\pm q$. The length of the plates is $l$, and their width coincides with the width of the tape. The tape is pulled out of the condenser with a force of $F$. Find the current through the resistance and the steady-state speed of the tape. The distance from the edge of the capacitor plates to the contact is much greater than the distance between them and much less than their length. | ||
| ### Solution | ### Solution | ||
| @@ -6,7 +6,7 @@Solution | |||
|  |  | ||
| Suppose the steady-state speed of the tape is $v$. Then, in a short time interval $\Delta t$ the charge $-qv\Delta t/l$ is removed from the region between the plates upward along the tape. Equivalently, conventional current $I=qv/l$ flows upward through the resistor, which generates heat at the rate of $I^2R=q^2v^2R/l^2$. Since this must equal |
Suppose the steady-state speed of the tape is $v$. Then, in a short time interval $\Delta t$ the charge $-qv\Delta t/l$ is removed from the region between the plates upward along the tape. Equivalently, conventional current $I=qv/l$ flows upward through the resistor, which generates heat at the rate of $I^2R=q^2v^2R/l^2$. Since this must equal the power input $Fv$ of the system, we have | ||
| \[\frac{q^2v^2R}{l^2}=Fv\qquad\Rightarrow\qquad v=\frac{Fl^2}{q^2R},\] | \[\frac{q^2v^2R}{l^2}=Fv\qquad\Rightarrow\qquad v=\frac{Fl^2}{q^2R},\] | ||
| and thus $I=Fl/(qR)$. | and thus $I=Fl/(qR)$. | ||
| #### Answer | #### Answer | ||
| $I=\frac{Fl}{qR}$, $v=\frac{Fl^2}{q^2R}$ | $I=\frac{Fl}{qR}$, $v=\frac{Fl^2}{q^2R}$ | ||
| unchanged lines 5 | |||