Question

Two large parallel metal plates are 6.0 cm apart. The magnitude of the electric field between them is 600 N/C. What is the potential difference between the plates? Note: the charge of an electron is 1.6 x 10-19 C.
Group of answer choices

40 V

18 V

21 V

36 V

Answer 1

To find the potential difference (V) between the two parallel plates, you can use the formula:

\[
V = E \times d
\]

where:
- \( V \) is the potential difference (in volts),
- \( E \) is the electric field strength (in N/C),
- \( d \) is the distance between the plates (in meters).

Given:
- \( E = 600 \, \text{N/C} \)
- \( d = 6.0 \, \text{cm} = 0.06 \, \text{m} \)

Now substitute the values into the formula:

\[
V = 600 \, \text{N/C} \times 0.06 \, \text{m} = 36 \, \text{V}
\]

Therefore, the potential difference between the plates is **36 V**.

The correct answer is:

**36 V**.