A parallel-plate capacitor is formed of of two 20.0 cm × 20.0 cm plates spaced 2.50 cm apart. The plates are charged to ±±4.00 nC. What is the magnitude of the electric field between the plates?
Question
A parallel-plate capacitor is formed of of two 20.0 cm × 20.0 cm plates spaced 2.50 cm apart. The plates are charged to ±±4.00 nC. What is the magnitude of the electric field between the plates?
Solution
The electric field (E) between the plates of a capacitor is given by the equation:
E = Q / (ε0 * A)
where:
- Q is the charge on each plate,
- ε0 is the permittivity of free space (approximately 8.85 x 10^-12 C^2/N*m^2), and
- A is the area of one plate.
Given:
- Q = 4.00 nC = 4.00 x 10^-9 C (since 1 nC = 10^-9 C),
- A = 20.0 cm x 20.0 cm = 400 cm^2 = 400 x 10^-4 m^2 (since 1 cm^2 = 10^-4 m^2).
Substituting these values into the equation gives:
E = (4.00 x 10^-9 C) / ((8.85 x 10^-12 C^2/N*m^2) * (400 x 10^-4 m^2)) = 1.13 x 10^6 N/C.
So, the magnitude of the electric field between the plates is approximately 1.13 x 10^6 N/C.
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