A particle leaves the origin with a speed of 2.4 times 10^6 m/s at 38 degrees to the positive x axis. It moves in a uniform electric field directed along positive y axis. Find Ey such that the particle will cross the x axis at x = 1.5 cm if the particle is an electron.

I don't know what I am doing wrong on this problem. So I used the equation y(t)= y0 + v0t + ((1/2)qEt^2)/m



I rearranged it to solve for E in terms of t and got E= 2.733e-5/t then I plugged in the t I got from finding x=vxt=1.5cm



My answer came out to be 3.44e3 N/C

I do not have time to run your numbers but your approach is correct.

constant Vx = 2.4*10^6 cos 38
so t = .015 /Vx

Vyi = 2.4*10^6 sin 38

y = Vyi t -(1/2)(qE/m) t^2 = 0 at x axis

To find the electric field Ey such that the particle will cross the x-axis at x = 1.5 cm, we can use the given information.

First, let's find the time it takes for the particle to cross the x-axis.
We have the initial velocity of the particle, v0 = 2.4 × 10^6 m/s, and the angle it makes with the positive x-axis, θ = 38°.

The x-component of the initial velocity can be found using the equation:
vx = v0 * cos(θ)
vx = 2.4 × 10^6 m/s * cos(38°)

Now, we can find the time it takes for the particle to cross the x-axis using the equation:
x = vxt
1.5 cm = (2.4 × 10^6 m/s * cos(38°)) * t

To convert the units of x from cm to meters, we need to multiply by 1 cm / 100 cm.
1.5 cm * (1/100) = (2.4 × 10^6 m/s * cos(38°)) * t

0.015 m = (2.4 × 10^6 m/s * cos(38°)) * t

Now, let's solve for t:
t = 0.015 m / (2.4 × 10^6 m/s * cos(38°))

Next, we can find the electric field Ey using the equation:
Ey = 2y / (t^2 * q / m)

Given that the particle is an electron, the charge q is -1.6 × 10^-19 C, and the mass m is 9.11 × 10^-31 kg.

Plugging in the values:

Ey = 2(0.015 m) / ((0.015 m / (2.4 × 10^6 m/s * cos(38°)))^2 * (-1.6 × 10^-19 C) / (9.11 × 10^-31 kg))

Ey ≈ 3.43 × 10^3 N/C

So a electric field Ey of approximately 3.43 × 10^3 N/C is needed for the electron to cross the x-axis at x = 1.5 cm.

To find the electric field Ey such that the particle will cross the x-axis at x = 1.5 cm, we need to use the equation of motion for a charged particle in a uniform electric field.

The equation you used, y(t) = y0 + v0t + ((1/2)qEt^2)/m, is correct. Let's go through the steps to derive the equation.

1. Start with the equation of motion in the y-direction:
y(t) = y0 + v0y t + (1/2) a y t^2

2. In this case, the initial position, y0, is 0 since the particle starts at the origin.

3. The initial velocity, v0y, can be found by resolving the initial velocity vector along the y-axis:
v0y = v0 sin θ, where v0 is the initial speed of the particle (2.4 * 10^6 m/s) and θ is the angle with the positive x-axis (38 degrees).

4. The acceleration in the y-direction, ay, is given by:
ay = (qE) / m, where q is the charge of the particle and m is the mass of the particle. In this case, considering the particle as an electron, q = -e (negative charge of the electron) and m is the mass of the electron.

5. Substitute the values for y0, v0y, and ay into the equation of motion:
y(t) = 0 + (2.4 * 10^6 m/s) * sin(38 degrees) * t + (1/2) * (-eE / me) * t^2

6. Simplify the equation:
y(t) = (2.4 * 10^6 m/s) * sin(38 degrees) * t - (1/2) * (eE / me) * t^2

Now, we want to find the electric field Ey such that the particle will cross the x-axis at x = 1.5 cm. This means that when y(t) = 0, x(t) = 0.015 m.

7. Solve for t when y(t) = 0:
0 = (2.4 * 10^6 m/s) * sin(38 degrees) * t - (1/2) * (eE / me) * t^2

8. Rearrange the equation to make it quadratic in t:
(1/2) * (eE / me) * t^2 = (2.4 * 10^6 m/s) * sin(38 degrees) * t

9. Simplify and divide both sides by t to isolate Ey:
(1/2) * (eE / me) * t = (2.4 * 10^6 m/s) * sin(38 degrees)
(1/2) * (eE / me) = (2.4 * 10^6 m/s) * sin(38 degrees) / t

10. Substitute the given x value for t (t = 0.015 m / (2.4 * 10^6 m/s * cos(38 degrees)):
(1/2) * (eE / me) = (2.4 * 10^6 m/s) * sin(38 degrees) / (0.015 m / (2.4 * 10^6 m/s * cos(38 degrees)))

11. Calculate the electric field Ey:
Ey = (2 * (2.4 * 10^6 m/s) * sin(38 degrees) * (2.4 * 10^6 m / s) * cos(38 degrees)) / (0.015 m * e / me)

Simplifying this expression will give you the value of Ey that satisfies the given conditions.