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

1 answer

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:

\[ \boxed{62.5 , \text{V}} \]