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.

Question

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.