Two $2\mathrm{.}4-cm-$diameter disks spaced $1\mathrm{.}6$ mm apart form a parallel$-$plate capacitor. The electric field between the disks is $4\mathrm{.}9{10}^5\frac{V}{m}.$

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Sumitht

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$$ Correct

MODEL: Energy is conserved. The electron's potential energy inside the capacitor can be found from the capacitor's electric potential.

SOLVE: The voltage across the capacitor is

$V_C=Ed=(4\mathrm{.}9{10}^5\frac{V}{m})(r_m)=784V$

Part B

An electron is launched from the negative plate. It strikes the positive plate at a speed $2\mathrm{.}2{10}^7\frac{m}{s}.$ What was the electron's speed as it left the negative plate?

Express your answer with the appropriate units.

View Available Hint$($s$)$

$v=$

Value