An electron in a vacuum chamber is fired with a speed of 8300 km/s toward a large, uniformly charged plate 75 cm away. the electron reaches a closest distance of 15 cm before being repelled. what is the plate’s surface charge density?

1.85*10^-27 C/m^2

Explanation: In order to explain this problem, firstly we have to consider the kinematic equations given by:

xf=vo*t-(a*t^2)/2 and we considerer xo=0 and xf= 15 cm the distance that the electron moves before changes its direction of movenet due to the influence of the electric field from the negative charged plate.

and

vf=vo-a*t  vf=0 when the electron starts to be repelled.

From these equations we have:

t=vo/a put it in the distante equation we have:

xf=vo*vo/a-(a/2)*(vo/a)^2 then xf=vo^2/a-(vo^2/2*a)= vo^2/(2*a)=

0.15 m=(8.3*10^6)^2/(2*a)    

we have used that: vo=8300 Km/s= 8300 km/s*1000 m/km=8.3*10^6m/s

a= (8.3*10^6)^2/(2*0.15)=5.74*10^13 m/s^2

Secondly, we use the second Newton law:

F=m*a where the force from the electric field is given by:

Fe=σ/εo  where σ is the  surface charge density and  the vaccum pernitivity is εo=8.85*10^-12 C^2/m^2*N

Finally, we have

Fe= σ/εo = m*a  

σ= m*a *εo= 9.1 *10^-31* 5.74*10^13 *8.85*10^-12=1.85*10^-27 C^2/m^2

the answer is c i did the same question so i know that's right!

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explanation: