To tackle this problem, we need to break it down into two parts: first, calculating the potential difference between the plates, and second, determining the speed of the electron as it exits the area between the plates. Let's dive into each part step by step.

Calculating the Potential Difference

The potential difference (V) between two parallel plates in an electric field can be calculated using the formula:

V = E × s

Where:

• V is the potential difference in volts (V).
• E is the electric field strength in newtons per coulomb (N/C).
• s is the separation between the plates in meters (m).

In this case, we have:

• E = 2.0 × 10⁴ N/C
• s = 1.0 cm = 0.01 m

Now, substituting the values into the formula:

V = (2.0 × 10⁴ N/C) × (0.01 m) = 2000 V

So, the potential difference between the plates is 2000 volts.

Finding the Speed of the Electron Upon Exiting

Next, we need to find the speed of the electron as it leaves the area between the plates. To do this, we can use the principle of energy conservation. The kinetic energy gained by the electron as it moves through the electric field is equal to the work done on it by the electric field.

The work done (W) on the electron can be expressed as:

W = q × V

Where:

• q is the charge of the electron (approximately -1.6 × 10^-19 C).
• V is the potential difference we just calculated (2000 V).

Now, substituting the values:

W = (-1.6 × 10^-19 C) × (2000 V) = -3.2 × 10^-16 J

Since we are interested in the kinetic energy gained, we can ignore the negative sign (as it indicates direction) and focus on the magnitude:

K.E. = W = 3.2 × 10^-16 J

The kinetic energy of the electron can also be expressed in terms of its mass (m) and speed (v) as:

K.E. = (1/2)mv²

The mass of an electron is approximately 9.11 × 10^-31 kg. Setting the two expressions for kinetic energy equal gives us:

3.2 × 10^-16 J = (1/2)(9.11 × 10^-31 kg)v²

Now, solving for v:

v² = (3.2 × 10^-16 J) / (0.5 × 9.11 × 10^-31 kg)

v² = (3.2 × 10^-16) / (4.555 × 10^-31)

v² ≈ 7.02 × 10¹⁴

v ≈ √(7.02 × 10¹⁴) ≈ 2.65 × 10⁷ m/s

Thus, the speed of the electron as it exits the area between the plates is approximately 2.65 × 10⁷ m/s.

Summary

In summary, we calculated the potential difference between the plates to be 2000 volts and found that the speed of the electron upon leaving the area between the plates is approximately 2.65 × 10⁷ m/s. This process illustrates the relationship between electric fields, potential difference, and the motion of charged particles.