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Two insulating plates are both uniformly charged in such a way that the potential difference between them is $V_2-V_1=20 \mathrm{~V}$. (i.e. plate 2 is at a higher potential). The plates are separated by $d=0.1 \mathrm{~m}$ and can be treated as infinitely large. An electron is released from rest on the inner surface of plate 1 . What is its speed when it hits plate $2 ?$
$\left(\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}, \mathrm{m}_{\mathrm{e}}=9.11 \times 10^{-31} \mathrm{~kg}\right)$
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$\left(\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}, \mathrm{m}_{\mathrm{e}}=9.11 \times 10^{-31} \mathrm{~kg}\right)$
Solution:
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Verified Answer
The correct answer is:
$2.65 \times 10^6 \mathrm{~m} / \mathrm{s}$
$2.65 \times 10^6 \mathrm{~m} / \mathrm{s}$
$\frac{1}{2} \mathrm{mv}^2=\mathrm{eV}$
$\mathrm{v}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}=2.65 \times 10^6 \mathrm{~m} / \mathrm{s}$
$\mathrm{v}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}=2.65 \times 10^6 \mathrm{~m} / \mathrm{s}$
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