An electron moving across capacitor problem

In summary: If you look at the problem, the electron is moving in the opposite direction of the electric field, so the electric field is acting against the motion of the electron. Thus, the acceleration should be negative. This will also give a positive value for 'a'. With these changes, you will get an answer of 0.00036 m for y.In summary, an electron is launched at a 45 degree angle and a speed of 5.0*10^6 m/s from the positive plate of a parallel-plate capacitor. It lands 4.0 cm away and the electric field inside the capacitor is -3.6x10^3 N/C. The minimum spacing between the plates is 0
  • #1
KillerZ
116
0

Homework Statement



An electron is launched at a 45 angle and a speed of 5.0*10^6 from the positive plate of the parallel-plate capacitor shown in the figure. The electron lands 4.0 cm away.

What is the electric field strength inside the capacitor?

What is the minimum spacing between the plates?

2wftmqc.jpg


Homework Equations



F = ma
Fon q = qE

The Attempt at a Solution



I am just wondering if this is correct?

Vo = 5x10^6 m/s

Voy = (sin45)(5x10^6 m/s) = 3.5x10^6 m/s

t = [tex]\frac{(0.04m)(\sqrt{2})}{5x10^6 m/s}[/tex] = 11x10^-9 s

0 = Voyt + (1/2)at^2

a = [tex]\frac{-(3.5x10^6 m/s)(11x10^-9 s)}{(\frac{1}{2})(11x10^-9 s)^2}[/tex]

= -6.4x10^-14 m/s^2

F=ma

=(9.11x10^-31 kg)(-6.4x10^14 m/s^2) = -5.8x10^-16 N

Fon q = qE

E = F/q = -5.8x10^-16 N / 1.60x10^-19 C

= -3.6x10^3 N/C

ymax = V0y/(2a)

= 3.5x10^6 m/s / [(2)(-6.4x10^-14 m/s^2)]

= 2x10^-9 m
 
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  • #2
How did you calculate t? The time of flight is dependent on the force acting on the electron, and the electric field is not known. Try setting up equations of motion eliminating t.

Namely,

[tex]y=y_0+v_{y0}t+\frac{1}{2}at^2[/tex]

[tex]x=v_{x0}t[/tex]

Solve for t in the second equation then substitute into the first. This gives two unknowns y and a. Find another kinematic equation that has y and a. This will give two equations with two unknowns. You will than be able to solve for y and a (hint: let y = ymax). Check your equation for ymax.
 
  • #3
Ok here is what I did:

[tex]y= y_0+v_{y0}t+\frac{1}{2}at^2[/tex]

[tex]x=v_{x0}t[/tex]

[tex]t=\frac{x}{v_{x0}}[/tex]

[tex]y= y_0+v_{y0}(\frac{x}{v_{x0}})+\frac{1}{2}a(\frac{x}{v_{x0}})^2[/tex]

[tex]v_{y}^2 = v_{y0}^2 + 2ay [/tex]

[tex]y=\frac{v_{y}^2 - v_{y0}^2}{2a}[/tex]

[tex]\frac{v_{y}^2 - v_{y0}^2}{2a}= y_0+v_{y0}(\frac{x}{v_{x0}})+\frac{1}{2}a(\frac{x}{v_{x0}})^2[/tex]

[tex]a^2=\frac{-v_{y0}^2 - v_{y0}(\frac{x}{v_{x0}})}{(\frac{x}{v_{x0}})^2}[/tex]

[tex]a^2=\frac{-(3.5x10^6 m/s)^2 - 3.5x10^6 m/s(\frac{0.04m}{3.5x10^6 m/s})}{(\frac{0.04m}{3.5x10^6 m/s})^2}[/tex]

= [tex]9.4x10^{28} m/s^2[/tex]

[tex]\sqrt{9.4x10^{28} m/s^2}[/tex] = [tex]3.1x10^{14} m/s^2[/tex]
 
Last edited:
  • #4
Your appoach is correct with the exception of two parts:

[tex]v_{y}^2 = v_{y0}^2 + 2ay [/tex]

[tex]y= y_0+v_{y0}t+\frac{1}{2}at^2[/tex]

Is the sign of 'a' positive or negative according to the coordinates? Another clue is evaluating

[tex]a^2=\frac{-(3.5x10^6 m/s)^2 - 3.5x10^6 m/s(\frac{0.04m}{3.5x10^6 m/s})}{(\frac{0.04m}{3.5x10^6 m/s})^2}[/tex]

Taking the square root of each side yields an imaginary number for 'a'.
 

Related to An electron moving across capacitor problem

1. What is a capacitor?

A capacitor is an electronic component used to store electrical energy. It consists of two conductive plates separated by an insulating material, known as a dielectric.

2. How does an electron move across a capacitor?

When a voltage is applied to a capacitor, one plate becomes positively charged and the other becomes negatively charged. This creates an electric field that causes electrons to move from one plate to the other, creating a flow of current.

3. What factors affect the movement of electrons across a capacitor?

The movement of electrons across a capacitor is affected by the voltage applied, the capacitance of the capacitor, and the resistance of the circuit.

4. What is the equation for calculating the charge on a capacitor?

The equation for calculating the charge on a capacitor is Q = CV, where Q is the charge (in coulombs), C is the capacitance (in farads), and V is the voltage (in volts).

5. How can I calculate the energy stored in a capacitor?

The energy stored in a capacitor can be calculated using the equation E = 1/2CV^2, where E is the energy (in joules), C is the capacitance (in farads), and V is the voltage (in volts).

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