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(C) Let $l = 3.0 \, cm = 3 	imes 10^{-2} \, m$ be the length of the plates and $d = 1.0 \, cm = 10^{-2} \, m$ be the separation between them.The time

Vedclass · Accepted Answer

$1.8 	imes 10^{11}$

Vedclass · Answer

(C) Let $l = 3.0 \, cm = 3 	imes 10^{-2} \, m$ be the length of the plates and $d = 1.0 \, cm = 10^{-2} \, m$ be the separation between them.The time $t$ taken by the electron to travel the length of the plates is given by $t = \frac{l}{v_x}$,where $v_x = 3 	imes 10^7 \, m/s$.$t = \frac{3 	imes 10^{-2} \, m}{3 	imes 10^7 \, m/s} = 10^{-9} \, s$.The electric field $E$ between the plates is $E = \frac{V}{d} = \frac{550 \, V}{10^{-2} \, m} = 5.5 	imes 10^4 \, V/m$.The force on the electron is $F = eE = ma_y$,so the acceleration $a_y = \frac{eE}{m} = \frac{e}{m} \cdot \frac{V}{d}$.The electron enters midway,so it needs to deflect by a vertical distance $y = \frac{d}{2}$ to strike the edge of the plate.Using the equation of motion $y = \frac{1}{2} a_y t^2$:$\frac{d}{2} = \frac{1}{2} \left( \frac{e}{m} \cdot \frac{V}{d} ight) t^2$.Rearranging for the specific charge $\frac{e}{m}$:$\frac{e}{m} = \frac{d^2}{V t^2} = \frac{(10^{-2})^2}{550 	imes (10^{-9})^2} = \frac{10^{-4}}{550 	imes 10^{-18}} = \frac{10^{14}}{550} \approx 1.818 	imes 10^{11} \, C/kg$.Thus,the specific charge is approximately $1.8 	imes 10^{11} \, C/kg$.

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