An electron is accelerated from rest by a uniform electric field of 3310V/M that is vertically pointing up, if the electron is traveling 6200000m/s how far (m) has it traveled in the electric field?

a=eE/m

s=v²/2a

To determine the distance traveled by the electron in the electric field, we can use the equation for accelerated motion:

d = (1/2) * a * t^2

Where:
d = distance traveled
a = acceleration
t = time

First, let's find the acceleration of the electron using the formula:

a = q * E / m

Where:
q = charge of the electron
E = electric field strength
m = mass of the electron

The charge of an electron, q, is constant and equal to the elementary charge, e, which is approximately -1.6 x 10^-19 C.

The electric field strength, E, is given as 3310 V/m.

The mass of an electron, m, is approximately 9.11 x 10^-31 kg.

Calculating the acceleration:

a = (-1.6 x 10^-19 C * 3310 V/m) / (9.11 x 10^-31 kg)
a ≈ -5.8 x 10^14 m/s^2 (Upward direction, as stated in the problem)

Now, we need to find the time it takes for the electron to reach a velocity of 6200000 m/s. We'll use the equation for uniformly accelerated motion:

v = u + a * t

Where:
v = final velocity
u = initial velocity
a = acceleration
t = time

The initial velocity, u, is given as 0 m/s since the electron is initially at rest.

Solving for time:

t = (v - u) / a
t = (6200000 m/s - 0 m/s) / (-5.8 x 10^14 m/s^2)
t ≈ -1.07 x 10^-11 s (Negative sign indicates upward direction)

Since time cannot be negative, we take the absolute value:

t ≈ 1.07 x 10^-11 s (Upward direction)

Finally, we can calculate the distance traveled using the first equation:

d = (1/2) * a * t^2
d = (1/2) * (-5.8 x 10^14 m/s^2) * (1.07 x 10^-11 s)^2
d ≈ 3.15 x 10^-5 m

Therefore, the electron has traveled approximately 3.15 x 10^-5 meters (or 31.5 μm) in the electric field.