Question

The electric field between two parallel plates has a magnitude of 180 N/C. The two plates are 2.5 cm apart. Recall that the magnitude of the charge on an electron is 1.602 × 10-19 C.

How much work is done to move an electron from the positive plate to the negative plate?

4.0 × 10-21 J
7.2 × 10-19 J
4.5 J
7200 J

Answer 1

To calculate the work done in moving an electron from the positive plate to the negative plate in an electric field, we can use the formula:

\[
W = q \cdot E \cdot d
\]

Where:
- \( W \) is the work done,
- \( q \) is the charge of the electron,
- \( E \) is the magnitude of the electric field, and
- \( d \) is the distance between the plates.

Given:
- \( q = 1.602 \times 10^{-19} \) C (the charge of the electron),
- \( E = 180 \) N/C (the electric field strength),
- \( d = 2.5 \) cm = 0.025 m (the distance between the plates).

Now substituting the values into the work formula:

\[
W = (1.602 \times 10^{-19} \, \text{C}) \cdot (180 \, \text{N/C}) \cdot (0.025 \, \text{m})
\]

Calculating it step by step:

1. Calculate \( 180 \times 0.025 \):
\[
180 \times 0.025 = 4.5
\]

2. Multiply by the charge of the electron:
\[
W = 1.602 \times 10^{-19} \cdot 4.5
\]

3. Calculate:
\[
W = 7.209 \times 10^{-19} \, \text{J}
\]

Since this result can be rounded to two significant figures, we get:

\[
W \approx 7.2 \times 10^{-19} \, \text{J}
\]

So the work done to move an electron from the positive plate to the negative plate is:

\[
\boxed{7.2 \times 10^{-19} \, \text{J}}
\]