In a parallel plate air capacitor,a cathode b | vedclass

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(B) The electric field $E$ between the plates is given by $E = \frac{\phi}{d}$.The force on an electron is $F = eE = \frac{e\phi}{d}$.The acceleration

Vedclass · Accepted Answer

$1.76$

Vedclass · Answer

(B) The electric field $E$ between the plates is given by $E = \frac{\phi}{d}$.The force on an electron is $F = eE = \frac{e\phi}{d}$.The acceleration of the electron is $a = \frac{F}{m} = \frac{e\phi}{md}$.The time taken by the electron to cross the plates of length $l$ is $t = \frac{l}{v_0}$.The vertical deflection $\Delta y$ is given by $\Delta y = \frac{1}{2}at^2 = \frac{1}{2} \left( \frac{e\phi}{md} ight) \left( \frac{l}{v_0} ight)^2$.Given values: $\phi = 400\ V$,$d = 2\ cm = 0.02\ m$,$l^2 = 100\ cm^2 \implies l = 10\ cm = 0.1\ m$,$v_0 = 10^8\ m/s$,$e = 1.6 	imes 10^{-19}\ C$,$m = 9.1 	imes 10^{-31}\ kg$.Substituting these values:$\Delta y = \frac{1}{2} 	imes \frac{1.6 	imes 10^{-19} 	imes 400}{9.1 	imes 10^{-31} 	imes 0.02} 	imes \left( \frac{0.1}{10^8} ight)^2$$\Delta y = \frac{1}{2} 	imes \frac{6.4 	imes 10^{-17}}{1.82 	imes 10^{-32}} 	imes 10^{-18} = \frac{6.4 	imes 10^{-35}}{3.64 	imes 10^{-32}} \approx 1.76 	imes 10^{-3}\ m = 1.76\ mm$.

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