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en/8.2.24.md
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| + | ### Statement | ||
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| + | $8.2.24.$ [Insert the problem statement] | ||
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| + | ### Solution | ||
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| + | 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 equal the power input $Fv$ of the system, we have | ||
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| + | \[\frac{q^2v^2R}{l^2}=Fv\qquad\Rightarrow\qquad v=\frac{Fl^2}{q^2R},\] | ||
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| + | and thus $I=Fl/(qR)$. | ||
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| + | #### Answer | ||
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| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $8.2.24.$ [Insert the problem statement] | |||
| ### 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 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 | |||
| [Insert a concise answer or boxed result] | |||