An electron is shot directly away from a uniformly charged plastic sheet Find σ.

In summary, the question involves finding the surface charge density of a nonconducting, flat, and very large plastic sheet. An electron is shot directly away from the sheet at a speed of 2.13 × 10^5 m/s. The graph of the electron's vertical velocity component v versus time t is given until the return to the launch point, with a velocity of 0 at 7 ps. Using the equations V = Vo + at, a = QE/m, and E = σ/ε₀, the surface charge density is found to be 1.523 microColoumbs/m^2. However, when the same calculations are done for a friend's problem, with an initial velocity of 2.
  • #1
nejibanana
12
1

Homework Statement



In figure (a) below, an electron is shot directly away from a uniformly charged plastic sheet, at speed vs = 2.13 × 10^5 m/s. The sheet is nonconducting, flat, and very large. Figure (b) below gives the electron's vertical velocity component v versus time t until the return to the launch point. What is the sheet's surface charge density?

There is also a graph that shows Velocity at various times. It is 0 at 7 ps.

Homework Equations



V = Vo + at

a = QE/m (Q = 1.609 x 10^-19, m = 9.109 x 10^-31)

E = σ/ε₀ (ε₀ = 8.85 x 10^-12) -> σ = Eε₀

The Attempt at a Solution



So I found acceleration with the first formula, then found E with the second, and found σ with the third. My answer was correct to the second decimal point, but when I did my friends' problems, the answer required me to subtract .4 from both of them to get the correct answer. I'm not sure if I did something wrong or if the homework website was wrong. If its any help, my solution was about 1.5. One of my friends problem had initial velocity of 2.03 x 10^5, all other numbers used were the same (give the same graph), and I calculated 1.4, but the correct answer was a little under 1.

Thanks for any help.
 
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  • #2


nejibanana said:

Homework Statement



In figure (a) below, an electron is shot directly away from a uniformly charged plastic sheet, at speed vs = 2.13 × 10^5 m/s. The sheet is nonconducting, flat, and very large. Figure (b) below gives the electron's vertical velocity component v versus time t until the return to the launch point. What is the sheet's surface charge density?

There is also a graph that shows Velocity at various times. It is 0 at 7 ps.

Homework Equations



V = Vo + at

a = QE/m (Q = 1.609 x 10^-19, m = 9.109 x 10^-31)

E = σ/ε₀ (ε₀ = 8.85 x 10^-12) -> σ = Eε₀

The Attempt at a Solution



So I found acceleration with the first formula, then found E with the second, and found σ with the third. My answer was correct to the second decimal point, but when I did my friends' problems, the answer required me to subtract .4 from both of them to get the correct answer. I'm not sure if I did something wrong or if the homework website was wrong. If its any help, my solution was about 1.5. One of my friends problem had initial velocity of 2.03 x 10^5, all other numbers used were the same (give the same graph), and I calculated 1.4, but the correct answer was a little under 1.

Thanks for any help.
Why not show us what you have done. Also, what units are you working in? An answer of 1.5 means nothing without units.

Get acceleration from [itex]\Delta v = a\Delta t[/itex] which it looks like you have done. What did you use for [itex]\Delta t[/itex]? The answer I get is:

a = 5.8e16 m/sec^-2

The force on the electron is F=ma where m is the mass of the electron. Since this is the coulomb force, qE, you have it correct that:

E = ma/q where q is the electron's charge.

I get: E = 9.11e-31 x 5.8e16/1.60e-19 = 3.3e5 N/coulomb

So [itex]\sigma = E\epsilon_0[/itex] = 3.3e5 x 8.85e-12 = 2.92e-6 coulomb/m^2

AM
 
  • #3


For change in time I used 7 picoseconds. Velocity changes from 2.13e5 to 0 in 7 ps. And I forget to edit it last night, but it wants the answers in microColoumbs/m^2.

So i got 2.13e5/ 7e-12 = a, which resulted in a = 3.04e16.

then I found E, so E = (9.109e-31) * (3.04e16) / (1.609e-19) . E = 172102.9.

Then σ = Eε₀. so σ = (172102.9) * ( 8.85 e-12) = 1.523e-6 C/m^2, converted to microColoumbs/m^2 is 1.523. Which was correct. 1.5313 is the answer given for my problem on the website.

For my friends I did, a = 2.03e5 / 7e-12 = 2.9e16

then E = (9.109e-31) * (2.9e16) / (1.609e-19). E = 164177.1287

then σ = (164177.1287) * (8.85 e -12) = 1.453e-6 C/m^2 = 1.453 microColoumbs/m^2. But according to the website the correct answer for his is 0.92875.

So I'm not sure what is causing the discrepancy between the problems of my 2 friends, since my problem worked out fine.

Thanks again.
 
  • #4


nejibanana said:
then σ = (164177.1287) * (8.85 e -12) = 1.453e-6 C/m^2 = 1.453 microColoumbs/m^2. But according to the website the correct answer for his is 0.92875.

That can't possibly be correct. 2.03 is just 5% lower than 2.13, so you'd expect the final answer to be 5% lower than 1.5. 0.929 is way off.
 

1. What is the significance of the variable σ in the context of an electron being shot away from a uniformly charged plastic sheet?

The variable σ represents the surface charge density of the plastic sheet, which is a measure of the amount of electric charge per unit area. It is used to calculate the electric field created by the sheet, which determines the trajectory of the electron.

2. How does the direction of the electric field affect the motion of the electron?

The direction of the electric field determines the direction of the force exerted on the electron. Since like charges repel each other, the electron will be pushed away from the sheet in a direction opposite to the electric field.

3. Can the value of σ change the path of the electron?

Yes, the value of σ directly affects the strength of the electric field, which in turn affects the acceleration of the electron. A higher value of σ will result in a stronger electric field and a more significant change in the path of the electron.

4. How is the surface charge density σ related to the total charge on the plastic sheet?

The surface charge density σ is equal to the total charge on the sheet divided by the area of the sheet. In other words, it represents the concentration of charge on the surface of the sheet.

5. How can the value of σ be determined experimentally?

The value of σ can be determined experimentally by measuring the electric field at various points around the plastic sheet and using the known relationship between electric field and surface charge density. This can be done using specialized equipment such as an electric field meter or a charged particle detector.

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