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Speed of proton parallel to insulating sheet

  • Thread starter GogumaDork
  • Start date
  • #1

Homework Statement


At time t=0, a proton is a distance of 0.360 meters from a very large insulating sheet of charge and is moving parallel to the sheet with speed 990 meters/second . The sheet has uniform surface charge density 2.34 * 10^-9 (C/M^2).

What is the speed of the proton at 7.10×10^−8 seconds?

Homework Equations


E = 1/(4pi(ε_0))*(|Q|/r^2)
q=σA
Charge of proton = 1.6*10^-19
F=ma=Eq

The Attempt at a Solution


I'm not really sure how to approach the problem as the textbook chapter examples only relate to solving problems relating to gauss's law, electric field, charge density and electric flux.

I tried assuming the question wants me to solve for E (electric field) using 1.6*10^-19C as the charge Q, and distance .36 meters. Then went back into F=ma = Eq and found a value for acceleration and plugged into V = v_0 + a(t) however the answer was incorrect.
 

Answers and Replies

  • #2
Curious3141
Homework Helper
2,843
87

Homework Statement


At time t=0, a proton is a distance of 0.360 meters from a very large insulating sheet of charge and is moving parallel to the sheet with speed 990 meters/second . The sheet has uniform surface charge density 2.34 * 10^-9 (C/M^2).

What is the speed of the proton at 7.10×10^−8 seconds?

Homework Equations


E = 1/(4pi(ε_0))*(|Q|/r^2)
q=σA
Charge of proton = 1.6*10^-19
F=ma=Eq

The Attempt at a Solution


I'm not really sure how to approach the problem as the textbook chapter examples only relate to solving problems relating to gauss's law, electric field, charge density and electric flux.

I tried assuming the question wants me to solve for E (electric field) using 1.6*10^-19C as the charge Q, and distance .36 meters. Then went back into F=ma = Eq and found a value for acceleration and plugged into V = v_0 + a(t) however the answer was incorrect.
This is a multistep problem. The first part is the toughest. You need to use calculus to work out the total instantaneous electric force on the proton. You can do this by envisioning the force on the proton as being contributed by infinitesimally thin annuli (rings) of charge on the sheet. The annuli spread out from a radius of zero to ∞ away from the proton. You need only consider the vertical component of the element of force, since the horizontal components all cancel out when you sum around each ring.

From this, you should be able to prove that the total (vertical) force on the proton at any one time is independent of its height above the sheet, and is given by [itex]F_y = \frac{e\sigma}{2\epsilon_0}[/itex]. That makes the calculation of the vertical acceleration easy, and you can use the usual kinematic equations for constant acceleration to figure out the final vertical component of velocity. The speed is given by the magnitude of the resultant of both horizontal and vertical velocities - so you need to do a vector sum. Use Pythagoras' theorem for this, it's easy.

There is another force on the proton - gravity, but it's tiny in comparison to the electric force, and can safely be neglected.

I get [itex]1.34 \times 10^{3} ms^{-1}[/itex] as the final answer.
 
Last edited:
  • #3
Curious3141
Homework Helper
2,843
87
A much quicker way to arrive at the result [itex]F_y = \frac{e\sigma}{2\epsilon_0}[/itex] is to use Gauss' law on a closed cylinder oriented with its long axis vertical (perpendicular to the sheet).
 

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