Question

The electric field between two parallel plates has a magnitude of 875 N/C. The positive plate is 0.002 m away from the negative plate.

What is the electric potential difference between the plates?

2.3 × 10-6 V
1.8 × 100 V
8.8 × 102 V
4.4 × 104 V

Answers

GPT-4o mini
The electric potential difference (\( V \)) between two parallel plates can be calculated using the formula:

\[
V = E \times d
\]

where:
- \( E \) is the magnitude of the electric field (in N/C),
- \( d \) is the distance between the plates (in meters).

Given:
- \( E = 875 \, \text{N/C} \)
- \( d = 0.002 \, \text{m} \)

Plugging in the values:

\[
V = 875 \, \text{N/C} \times 0.002 \, \text{m} = 1.75 \, \text{V}
\]

Now, we can convert this result into standard scientific notation:

\[
1.75 \, \text{V} = 1.75 \times 10^0 \, \text{V}
\]

None of the options you provided match this value. It appears there might be a misunderstanding or a miscalculation of the units or values you are considering.

Let's ensure the original data is correct. The electric potential difference across the plates calculated is indeed 1.75 V. If you have any other parameters or details to consider, please provide them for further evaluation.

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