Motion of point charges in electronic fields

[timtng]

An electron starts at the position (0,0) with an initial velocity 5,000,000 m/s at 45 degrees angle to the x axis. THe electric field is in the positive y direction and has a magnitude of 35000 N/C. At what location will the electron strikes?

Please help me setup this problem.

 

[Gza]

You really just have to think of this the same way you think of a projectile motion problem (an object acting under the influence of gravity.) The force on the electron is always down within the uniform electric field. This means the electron will strike somewhere on the x-axis. Since you know the charge, and the electric field it's in, you know the acceleration
(use F=qE and F=ma). Since you know the acceleration (and it's constant) use some kinematical relations and solve the problem.

 

[Graeme Robinson]

To solve this problem, we can use the equations of motion for a point charge in an electric field. The first equation we can use is the acceleration equation, which states that the acceleration of a point charge is equal to the electric field strength divided by the charge of the particle. In this case, the charge of the electron is negative and has a magnitude of 1.6 x 10^-19 C. So, the acceleration of the electron will be:

a = E/q = (35000 N/C)/(1.6 x 10^-19 C) = 2.1875 x 10^23 m/s^2

Next, we can use the kinematic equation for position, which states that the final position (xf) is equal to the initial position (xi) plus the initial velocity (vi) multiplied by time (t) plus half the acceleration (a) multiplied by the square of time (t^2). In this case, we want to find the final position (xf) when the electron strikes, so we can rearrange the equation to solve for xf:

xf = xi + vi*t + (1/2)*a*t^2

Since the electron starts at position (0,0) and has an initial velocity of 5000000 m/s at a 45 degree angle, we can break down the initial velocity into its x and y components using trigonometry:

vx = vi*cos(45) = 5000000*cos(45) = 3535534.3 m/s
vy = vi*sin(45) = 5000000*sin(45) = 3535534.3 m/s

Now, we can plug in these values into the equation for xf:

xf = 0 + 3535534.3*t + (1/2)*2.1875 x 10^23*t^2

Since we want to find the time when the electron strikes, we can set xf equal to the distance the electron travels in the y direction, which is the height of the electric field (yf). So, we have:

yf = 3535534.3*t + (1/2)*2.1875 x 10^23*t^2

To solve for t, we can use the quadratic formula:

t = (-b ± √(b^2 - 4ac))/2a

Plugging in the values for a, b,