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Question

(A) The electric field due to an infinite plane sheet is $E = \frac{\sigma}{2 \epsilon_0}$.Since the electron is negatively charged,the force on it is

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) The electric field due to an infinite plane sheet is $E = \frac{\sigma}{2 \epsilon_0}$.Since the electron is negatively charged,the force on it is $F = -eE = -\frac{e \sigma}{2 \epsilon_0}$.The acceleration of the electron is $a = \frac{F}{m} = -\frac{\sigma e}{2 \epsilon_0 m}$.Given initial velocity $u = 1 \,m/s$,time $t = 1 \,s$,and displacement $S = -1 \,m$ (towards the sheet).Using the equation of motion $S = ut + \frac{1}{2} at^2$:$-1 = (1)(1) + \frac{1}{2} \left( -\frac{\sigma e}{2 \epsilon_0 m} \right) (1)^2$.$-1 = 1 - \frac{\sigma e}{4 \epsilon_0 m}$.$2 = \frac{\sigma e}{4 \epsilon_0 m}$.$\sigma = 8 \left[ \frac{m \epsilon_0}{e} \right]$.Comparing with $\alpha \left[ \frac{m \epsilon_0}{e} \right]$,we get $\alpha = 8$.

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