A capacitor with surface charge density sigma | vedclass

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Question

(B) For the electron to pass without deflection,the electric force must be balanced by the magnetic force: $|\overrightarrow{F}_e| = |\overrightarrow{

Vedclass · Accepted Answer

$\frac{\varepsilon_0 l B}{\sigma}$

Vedclass · Answer

(B) For the electron to pass without deflection,the electric force must be balanced by the magnetic force: $|\overrightarrow{F}_e| = |\overrightarrow{F}_m|$.Since $|\overrightarrow{F}_e| = eE$ and $|\overrightarrow{F}_m| = evB$,we have $eE = evB$,which implies $v = \frac{E}{B}$.The electric field between the plates of a capacitor is given by $E = \frac{\sigma}{\varepsilon_0}$.Substituting this into the velocity equation,we get $v = \frac{\sigma}{\varepsilon_0 B}$.The time $t$ taken to travel the length $l$ is $t = \frac{l}{v}$.Substituting the value of $v$,we get $t = \frac{l}{(\sigma / \varepsilon_0 B)} = \frac{\varepsilon_0 l B}{\sigma}$.

$A$ capacitor with surface charge density $\sigma$ has a magnetic field $\overrightarrow{B}$ between its two plates. If an electron passes through without any deflection,how long will it take to exit the plates of length $l$?

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