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(C) The force on the electron due to the electric field $E$ is $F_E = eE$.The force on the electron due to the magnetic field $B$ is $F_B = evB$.For t

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The force on the electron due to the electric field $E$ is $F_E = eE$.The force on the electron due to the magnetic field $B$ is $F_B = evB$.For the electron to move undeflected,these two forces must be equal and opposite:$eE = evB \implies v = \frac{E}{B}$.The electric field between the plates of a capacitor is given by $E = \frac{\sigma}{\varepsilon_0}$.Substituting this into the velocity expression,we get $v = \frac{\sigma}{\varepsilon_0 B}$.The time $t$ taken to travel a distance $\ell$ is $t = \frac{\ell}{v}$.Substituting the value of $v$,we get $t = \frac{\ell}{\frac{\sigma}{\varepsilon_0 B}} = \frac{\varepsilon_0 \ell B}{\sigma}$.

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