An air-filled parallel-plate capacitor has plates of area
2.30 cm2 separated by 1.50 mm. (a) Find the value of its
capacitance. The capacitor is connected to a 12.0-V battery.
(b) What is the charge on the capacitor? (c) What is the
magnitude of the uniform electric field between the plates?
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To find the answers to these questions, we'll need to use the formulas related to capacitance, charge, and electric field.
(a) The capacitance of a parallel-plate capacitor is given by the formula:
C = ε₀ * (A / d)
where C is the capacitance, ε₀ is the permittivity of free space (ε₀ ≈ 8.85 x 10⁻¹² F/m), A is the area of the plates, and d is the separation distance between the plates.
Plugging in the values given:
A = 2.30 cm² = 2.30 x 10⁻⁴ m²
d = 1.50 mm = 1.50 x 10⁻³ m
Calculating:
C = (8.85 x 10⁻¹² F/m) * (2.30 x 10⁻⁴ m² / 1.50 x 10⁻³ m)
C ≈ 36.20 x 10⁻¹² F
C ≈ 36.20 pF
So, the capacitance of the parallel-plate capacitor is approximately 36.20 picofarads.
(b) The charge on a capacitor is given by the formula:
Q = C * V
where Q is the charge, C is the capacitance, and V is the voltage across the capacitor.
Plugging in the values given:
C = 36.20 x 10⁻¹² F
V = 12.0 V
Calculating:
Q = (36.20 x 10⁻¹² F) * (12.0 V)
Q ≈ 434.4 x 10⁻¹² C
Q ≈ 434.4 picocoulombs
So, the charge on the capacitor is approximately 434.4 picocoulombs.
(c) The magnitude of the uniform electric field between the plates of a parallel-plate capacitor is given by the formula:
E = V / d
where E is the electric field, V is the voltage across the capacitor, and d is the separation distance between the plates.
Plugging in the values given:
V = 12.0 V
d = 1.50 mm = 1.50 x 10⁻³ m
Calculating:
E = (12.0 V) / (1.50 x 10⁻³ m)
E ≈ 8000 V/m
So, the magnitude of the uniform electric field between the plates is approximately 8000 volts per meter.