An electron is accelerated in the uniform field E = 2.0 x 104 N/C between two parallel charged plates (Negative charged plate is on the left and positively charges plate is on the right). The separation of the plates is 1.5 cm. The electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate. With what speed does it leave the hole?

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To find the speed at which the electron leaves the hole, we can use the equations of motion. The first step is to determine the acceleration of the electron in the electric field between the plates.

Given:
Electric field (E) = 2.0 x 10^4 N/C
Separation of plates (d) = 1.5 cm = 0.015 m

We know that the electric force experienced by an electron can be given by the equation:

F = q * E

Where F is the force, q is the charge of the particle, and E is the electric field.

The electric force acting on the electron can be written as the product of the electron's charge (q = -e, where e is the elementary charge) and the electric field:

F = -e * E

Since force (F) is equal to mass (m) multiplied by acceleration (a), we can write:

ma = -eE

Rearranging the equation to isolate acceleration:

a = (-eE) / m

The negative sign indicates that the acceleration is opposite to the direction of the electric field, which makes sense as the electron is negatively charged, and the electric field is directed from positive to negative.

Now, we can calculate the acceleration by substituting the known values:

a = (-1.6 x 10^-19 C) * (2.0 x 10^4 N/C) / (9.11 x 10^-31 kg)

After evaluating the expression, we get the acceleration of the electron.

Next, we can use the equation of motion:

v^2 = u^2 + 2as

Where:
v = final velocity (the speed at which the electron leaves the hole)
u = initial velocity (which is 0 m/s since the electron starts from rest)
a = acceleration (calculated in the previous step)
s = displacement (which is the separation between the plates, 0.015 m)

Rearranging the equation to solve for v:

v^2 = 0^2 + 2 * a * s

v^2 = 2 * a * s

Taking the square root of both sides, we get:

v = √(2 * a * s)

Now, we can substitute the values of acceleration and displacement:

v = √(2 * a * s) = √(2 * (-1.6 x 10^-19 C) * (2.0 x 10^4 N/C) * (0.015 m) / (9.11 x 10^-31 kg))

Evaluating this expression will give us the speed at which the electron leaves the hole.

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