The plates described in Problem 25.27 are in vacuum. An electron \( \left(q=-e, m_{e}=9.1 \times 10^{-31} \mathrm{~kg}\right) \) is released at the negative plate and falls freely to the positive plate. How fast is it going just before it strikes the plate? Ans. \( \quad 7.3 \times 10^{5} \mathrm{~m} / \mathrm{s} \)
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The electron is released at the negative plate, which means it starts with zero initial velocity. Show more…
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The plates described in Problem $25.29$ are in vacuum. An electron $\left(q=-e, m_{e}=9.1 \times 10^{-31} \mathrm{~kg}\right)$ is released at the negative plate and falls freely to the positive plate. How fast is it going just before it strikes the plate?
Two charged metal plates in vacuum are $15 \mathrm{~cm}$ apart as drawn in Fig. 24-7. The electric field between the plates is uniform and has a strength of $E=3000 \mathrm{~N} / \mathrm{C}$. An electron $\left(q=-e, m_{e}=9.1 \times 10^{-31}\right.$ $\mathrm{kg}$ ) is released from rest at point $P$ just outside the negative plate. (a) How long will it take to reach the other plate? (b) How fast will it be going just before it hits? The electric field lines show the force on a positive charge. (A positive charge would be repelled to the right by the positive plate and attracted to the right by the negative plate.) An electron, being negative, will experience a force in the opposite direction, toward the left, of magnitude $$ F_{E}=|q| E=\left(1.6 \times 10^{-19} \mathrm{C}\right)(3000 \mathrm{~N} / \mathrm{C})=4.8 \times 10^{-16} \mathrm{~N} $$ Because of this force, the electron experiences an acceleration toward the left given by $$ a=\frac{F_{E}}{m}=\frac{4.8 \times 10^{-16} \mathrm{~N}}{9.1 \times 10^{-31} \mathrm{~kg}}=5.3 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2} $$ In the motion problem for the electron released at the negative plate and traveling to the positive plate, $$ v_{i}=0 x=0.15 \mathrm{~m} a=5.3 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2} $$ (a) Fiom $x=v_{i} t+\frac{a}{a}$ $$ t=\sqrt{\frac{2 x}{a}}=\sqrt{\frac{(2)(0.15 \mathrm{~m})}{5.3 \times 10^{4^{4}} \mathrm{~m} / \mathrm{s}^{2}}}=2.4 \times 10^{-8} \mathrm{~s} $$ (b) $v=v_{i}+a t=0+\left(5.3 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2}\right)\left(2.4 \times 10^{-8} \mathrm{~s}\right)=1.30 \times 10^{7}$ $\mathrm{m} / \mathrm{s}$ As you will see in Chapter 41, relativistic effects begin to become important at speeds above this. Therefore, this approach must be modified for very fast particles.
Two charged metal plates in vacuum are $15 \mathrm{~cm}$ apart as drawn in Fig. $24-7$. The electric field between the plates is uniform and has a strength of $E=3000 \mathrm{~N} / \mathrm{C}$. An electron $\left(q=-e, m_{e}=9.1 \times 10^{-31} \mathrm{~kg}\right)$ is released from rest at point $P$ just outside the negative plate. ( $a$ ) How long will it take to reach the other plate? $(b)$ How fast will it be going just before it hits? The electric field lines show the force on a positive charge. (A positive charge would be repelled to the right by the positive plate and attracted to the right by the negative plate.) An electron, being negative, will experience a force in the opposite direction, toward the left, of magnitude $$F_{E}=|q| E=\left(1.6 \times 10^{-19} \mathrm{C}\right)(3000 \mathrm{~N} / \mathrm{C})=4.8 \times 10^{-16} \mathrm{~N}$$ Because of this force, the electron experiences an acceleration toward the left given by $$a=\frac{F_{E}}{m}=\frac{4.8 \times 10^{-16} \mathrm{~N}}{9.1 \times 10^{-31} \mathrm{~kg}}=5.3 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2}$$ In the motion problem for the electron released at the negative plate and traveling to the positive plate, $$v_{i}=0 \quad x=0.15 \mathrm{~m} \quad a=5.3 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2}$$ (a) From $x=v_{i} t+\frac{1}{2} a t^{2}$ we have $$t=\sqrt{\frac{2 x}{a}}=\sqrt{\frac{(2)(0.15 \mathrm{~m})}{5.3 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2}}}=2.4 \times 10^{-8} \mathrm{~s}$$ (b) $\quad v=v_{i}+a t=0+\left(5.3 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2}\right)\left(2.4 \times 10^{-8} \mathrm{~s}\right)=1.30 \times 10^{7} \mathrm{~m} / \mathrm{s}$ As you will see in Chapter 41 , relativistic effects begin to become important at speeds above this. Therefore, this approach must be modified for very fast particles.
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