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

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards