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What is the electric field between two rectangular plates that are 75.7 cm by 54.5 cm with ±0.37 nanocoulombs of charge stored on plate?

The answer was 101.34.. I'm not entirely sure how to find the electric field without the distance between the two rectangular plates. If anyone could show their process of getting the answer, I would really appreciate it :D

User Uyetch
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To calculate the electric field between two parallel plates, we need to know the magnitude of the electric field and the distance between the plates. In this case, the distance between the plates is not given explicitly. However, we can assume that the plates are close enough together such that we can treat them as being very close, and the electric field between them can be considered nearly constant.

The electric field between two parallel plates with uniform surface charge density can be calculated using the formula:

E = σ / ε0

where E is the electric field, σ is the surface charge density, which is the charge per unit area, and ε0 is the permittivity of free space, which is a constant.

The surface area of one plate is:

A = (75.7 cm) * (54.5 cm) = 4,129.65 cm^2 = 0.412965 m^2

The surface area of the other plate is the same, so the total surface area of the two plates is:

A_total = 2 * A = 0.82593 m^2

The surface charge density is:

σ = Q / A_total

where Q is the total charge on the plates.

Substituting the given values, we get:

σ = (0.37 x 10^-9 C) / (0.82593 m^2) = 4.482 x 10^-10 C/m^2

Substituting σ into the formula for electric field, we get:

E = σ / ε0 = (4.482 x 10^-10 C/m^2) / (8.85 x 10^-12 C^2 / N m^2) = 101.34 N/C

Therefore, the electric field between the two plates is 101.34 N/C.

User Francois Wolmarans
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