An electron is projected towards a metal plat | vedclass

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(D) The electric field $E$ near a charged conducting plate is given by $E = \frac{|\sigma|}{\in_0}$.The force on the electron is $F = eE = e \frac{|\s

Vedclass · Accepted Answer

$\frac{E_k \in_0}{e\sigma}$

Vedclass · Answer

(D) The electric field $E$ near a charged conducting plate is given by $E = \frac{|\sigma|}{\in_0}$.The force on the electron is $F = eE = e \frac{|\sigma|}{\in_0}$.Since the electron is moving against the electric field (repelled by the negative plate),the work done by the electric field is negative,or the work done by the external kinetic energy to reach the plate is equal to the change in potential energy.Using the work-energy theorem,the initial kinetic energy $E_k$ is completely converted into work done against the electric force over distance $d$:$E_k = F \cdot d = (eE) \cdot d$Substituting $E = \frac{|\sigma|}{\in_0}$:$E_k = e \left( \frac{|\sigma|}{\in_0} \right) d$Solving for $d$:$d = \frac{E_k \in_0}{e|\sigma|}$Given the magnitude of charge density is $|\sigma|$,the expression matches option $D$.

An electron is projected towards a metal plate with a surface charge density $\sigma = -2 \times 10^{-6} \ C/m^2$. If the electron strikes the plate with zero velocity,calculate the distance $d$ between the electron gun and the plate in terms of its initial kinetic energy $E_k$.

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