Question
The electric field between two parallel plates has a magnitude of 1250 N/C. The positive plate is 0.05 m away from the negative plate.
what is the electric potential difference between the plates, rounded to the tenths place
what is the electric potential difference between the plates, rounded to the tenths place
Answers
GPT-4o mini
The electric potential difference \( V \) between two parallel plates can be calculated using the formula:
\[
V = E \cdot d
\]
where:
- \( E \) is the magnitude of the electric field,
- \( d \) is the distance between the plates.
Given:
- \( E = 1250 \, \text{N/C} \)
- \( d = 0.05 \, \text{m} \)
Substituting the values into the formula:
\[
V = 1250 \, \text{N/C} \cdot 0.05 \, \text{m} = 62.5 \, \text{V}
\]
Rounding to the tenths place, the electric potential difference is:
\[
\boxed{62.5 \, \text{V}}
\]
\[
V = E \cdot d
\]
where:
- \( E \) is the magnitude of the electric field,
- \( d \) is the distance between the plates.
Given:
- \( E = 1250 \, \text{N/C} \)
- \( d = 0.05 \, \text{m} \)
Substituting the values into the formula:
\[
V = 1250 \, \text{N/C} \cdot 0.05 \, \text{m} = 62.5 \, \text{V}
\]
Rounding to the tenths place, the electric potential difference is:
\[
\boxed{62.5 \, \text{V}}
\]