Question

Two large parallel metal plates are 3.0 cm apart. The magnitude of the electric field between them is 700 N/C. What work is done when one electron is moved from the positive to the negative plate? Note: the charge of an electron is 1.6 x 10^-19 C.
Group of answer choices

1.6 x 10^-19 J

3.36 x 10^-18 J

4.98 x 10^-17 J

5.36 x 10^-20 J

Answer 1

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

\[
W = q \cdot V
\]

Where:
- \( W \) is the work done,
- \( q \) is the charge of the electron, and
- \( V \) is the potential difference between the plates.

First, we need to find the potential difference \( V \). The potential difference can be related to the electric field \( E \) and the distance \( d \) between the plates using the formula:

\[
V = E \cdot d
\]

Given:
- \( E = 700 \, \text{N/C} \)
- \( d = 3.0 \, \text{cm} = 0.03 \, \text{m} \)

Calculating \( V \):

\[
V = 700 \, \text{N/C} \times 0.03 \, \text{m} = 21 \, \text{V}
\]

Next, we can find the work done when moving an electron (with charge \( q = 1.6 \times 10^{-19} \, \text{C} \)):

\[
W = q \cdot V = (1.6 \times 10^{-19} \, \text{C}) \times (21 \, \text{V})
\]

Calculating \( W \):

\[
W = 1.6 \times 10^{-19} \times 21 = 3.36 \times 10^{-18} \, \text{J}
\]

Thus, the work done when the electron is moved from the positive plate to the negative plate is:

\[
\boxed{3.36 \times 10^{-18} \, \text{J}}
\]