What is the electric field strength inside the capacitor?

[YamiBustamante]

Homework Statement

[/B]
An electron is launched at a 45∘ angle and a speed of 5.0×10^6 m/s from the positive plate of the parallel-plate capacitor shown in the figure (Figure 1) . The electron lands 4.0 cm away.

a)
What is the electric field strength inside the capacitor?

b)
What is the smallest possible spacing between the plates?

Homework Equations

F = qE
Equations used for Projectile Motion

The Attempt at a Solution

For Part A[/B]
Okay so first I thought that I may need the acceleration in the y-direction to use F = qE, so I separated the velocity into separate components.
Vx = 5.0×10^6 m/s * cos(45)
Vy = 5.0×10^6 m/s * sin(45)
Then I found the time.
0.04m = (5.0×10^6 m/s * cos(45))*t
t = 1.131*10^-8 seconds
Then I proceeded to finding the acceleration with the time I found by using the highest point reached (0 m/s)
0 m/s = 5.0×10^6 m/s * sin(45) + a (1.131*10^-8 sec)
a = -2.69 * 10^4 m/s^2

After that I found the force on the electron using its mass, 9.10938356 × 10^-31 kilograms
F = (9.10938356 × 10^-31 kg)*(-2.69 * 10^4 m/s^2)
F= -2.459*10^-16 N
Then used that to find E in F= qE using the charge of the electron 1.6*10^-19 C
2.459*10^-16 N = (1.6*10^-19 C)*(E)
E = -1536.57 N/C

But it's wrong. Can someone tell me what I did wrong?For Part B.
I tried calculating the amount of distance the electron traveled in the vertical direction.
0^2 = (5.0×10^6 m/s * sin(45))^2+2(-2.69 * 10^4 m/s^2)*y
y= 0.023 m

But it's wrong. I think I might have misunderstood the question. How do I find the answer?Any help is much appreciated!

 

[TSny]

Does the electron reach max height at the same time that Δx = 4.0 cm?

 

[gneill]

A possibly helpful hint (or approach to try): Consider the range equation for projectile motion.

 

[collinsmark]

Homework Statement

[/B]
An electron is launched at a 45∘ angle and a speed of 5.0×10^6 m/s from the positive plate of the parallel-plate capacitor shown in the figure (Figure 1) . The electron lands 4.0 cm away.

a)
What is the electric field strength inside the capacitor?

b)
What is the smallest possible spacing between the plates?

Homework Equations

F = qE
Equations used for Projectile Motion

The Attempt at a Solution

For Part A[/B]
Okay so first I thought that I may need the acceleration in the y-direction to use F = qE, so I separated the velocity into separate components.
Vx = 5.0×10^6 m/s * cos(45)
Vy = 5.0×10^6 m/s * sin(45)
Then I found the time.
0.04m = (5.0×10^6 m/s * cos(45))*t
t = 1.131*10^-8 seconds

Take a moment here to reflect on what total time [itex] t [/itex] represents. Is it the time it takes for the electron to reach the highest point in its motion or the time it takes the electron to traverse the entire 4 cm?

Then I proceeded to finding the acceleration with the time I found by using the highest point reached (0 m/s)
0 m/s = 5.0×10^6 m/s * sin(45) + a (1.131*10^-8 sec)

Take a look at the value of [itex] t [/itex] that you are using in your [itex] 0 = v_0 + at [/itex] equation. Should the value of [itex] t [/itex] used here be the time it takes for the electron to reach its highest point or the time it takes for the electron to traverse a full 4 cm?

When the electron reaches its highest point, how far has it traveled horizontally?

 

[hcrawford8]

7 years later and I have to solve this exact problem for my HW, so I am sure there will be someone else looking here for help. Haven't figured out the second part yet but I was able to solve the first part of the problem.
So, you want to start by finding the time it takes from the moment the electron is launch to the point where it lands. This can be found by taking V=d/t and rearranging to solve for the time. However since the electron took the same time to travel in the Y direction as it did in the X direction, you have to use the initial velocity [V][/0] given in the problem. --> t=0.04m/(5*10^6 m/s) = 8*10^-9s

After finding t, you can now take the kinematic equation for distance and use it to solve for acceleration.
d=[V][/0]t+(1/2)at^2 --> a=(2d-[V][/0]t)/(t^2) = (2*0.04m-5*10^-6m/s * 8*10^-9s) / (8*10^-9s)^2 = 6.25*10^14m/s^2

From here he can find the electric field by using a=eE/m, where 'e' is the charge of the electron and 'm' is the mass of the electron. Rearranging the equation to solve for the electric field. --> E=ma/e
E=(9.11*10^-31kg)(6.25*10^14m/s^2) / (1.6*10^-19C) = 3550N/C

 

[nasu]

You have to use the x component of the velocity and not the total velocity to find the time it takes to travel 4 cm.

 

[kuruman]

@nasu's point is well taken. More directly, since the electron returns to the same height from which it was launched, you can also use the range equation for projectile motion to find the acceleration $$R=\frac{2v_{0x}v_{0y}}{a}.$$