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

1 answer

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:

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