How far should an electron be fired so it stops before reaching the plate?

In summary, the conversation is about solving a physics problem involving mass, charge, velocity, force, electric field, potential difference, and displacement. The correct answer is expected to be 0.001475m, but the individual is getting a different answer. They have tried solving it using both energy and velocity, but both methods result in the same incorrect answer. The conversation ends with a question about the expected answer.
  • #1
Fontseeker
30
4
Homework Statement
In the figure below an electron is shot directly toward the center of a large metal plate that has surface charge density -1.20 × 10^-6 C/m2. If the initial kinetic energy of the electron is 1.60 × 10^-17 J and if the electron is to stop (due to electrostatic repulsion from the plate) just as it reaches the plate, how far from the plate must the launch point be?
Relevant Equations
F=ma
E=(1/2)mV^2
I am trying to solve the following problem:

Screen Shot 2019-09-09 at 9.07.22 PM.png

My attempt at the problem:

$$m = 9.11 * 10^{-31}kg$$
$$q = 1.6 * 10^{-19} C$$

$$V_0 = \sqrt{\frac{2KE}{m}} = \sqrt{\frac{2 * 1.6 * 10^{-17}}{1.6 * 10^{-19}}} = 5,926,739 m/s$$
$$F_{net} = -F_{e} = ma$$
$$a = \frac{-F_{e}}{m} = ma$$
$$E = \frac{\phi}{2\epsilon_0}$$
$$F_e = \frac{E}{q} = \frac{\phi}{2\epsilon_0q}$$
$$a = -\frac{q\phi}{2m\epsilon_0} = \frac{1.6 * 10^{-19} * -1.20 * 10^{-6}}{2 * 9.11 * 10^{-31} * 8.85 * 10^{-12}} = -1.191 * 10^{16} m/s^2$$
$$V_f^2 = V_0^2 + 2ad$$
$$V_f = 0$$
$$-2ad = V_0^2$$
$$d = \frac{V_0^2}{2a} = -\frac{5,926,739}{2 * -1.191 * 10^{16}} = 0.001475m$$

I got that the answer is 0.001475m, but this is not correct. Does anyone know what I am doing wrong?
 
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  • #2
What is ##\frac{10^{-17}}{10^{-19}}##?
But you are making work for yourself in finding the velocity. Just deal with energy.
 
  • #3
haruspex said:
What is ##\frac{10^{-17}}{10^{-19}}##?
But you are making work for yourself in finding the velocity. Just deal with energy.
that's a typo, should be:
$$\sqrt{\frac{2 * 1.6 * 10^{-17}}{9.11 * 10^{-31}}} = 5,936,739$$

I've dealt with energy and I get the same thing. Either way, it should work with velocity.
 
  • #4
Fontseeker said:
that's a typo, should be:
$$\sqrt{\frac{2 * 1.6 * 10^{-17}}{9.11 * 10^{-31}}} = 5,936,739$$

I've dealt with energy and I get the same thing. Either way, it should work with velocity.
Yes, I'm getting the same answer.
Do you know what the supposed answer is?
 

Related to How far should an electron be fired so it stops before reaching the plate?

1. How is the distance between the electron and the plate determined?

The distance between the electron and the plate is determined by the strength of the electric field and the initial velocity of the electron.

2. What factors affect the distance an electron travels before reaching the plate?

The distance an electron travels before reaching the plate is affected by the strength of the electric field, the initial velocity of the electron, and the mass of the electron.

3. Is there a formula to calculate the distance an electron will travel before reaching the plate?

Yes, the distance an electron will travel before reaching the plate can be calculated using the equation d = (v^2 * sin(2θ))/g, where v is the initial velocity, θ is the angle of firing, and g is the acceleration due to gravity.

4. How does changing the strength of the electric field affect the distance an electron travels before reaching the plate?

Increasing the strength of the electric field will increase the force acting on the electron, causing it to accelerate faster and cover a greater distance before reaching the plate.

5. Can an electron be fired from any distance and still be stopped before reaching the plate?

No, the distance an electron can travel before reaching the plate is limited by the strength of the electric field and the initial velocity of the electron. If these factors are too high, the electron may not be able to be stopped before reaching the plate.

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