The potential difference across the parallel plate capacitor is 2.26 millivolts
<h3>Capacitance of a parallel plate capacitor</h3>
The capacitance of the parallel plate capacitor is given by C = ε₀A/d where
- ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m,
- A = area of plates and
- d = distance between plates = 4.0 mm = 4.0 × 10⁻³ m.
<h3>Charge on plates</h3>
Also, the surface charge on the capacitor Q = σA where
- σ = charge density = 5.0 pC/m² = 5.0 × 10⁻¹² C/m² and
- a = area of plates.
<h3>
The potential difference across the parallel plate capacitor</h3>
The potential difference across the parallel plate capacitor is V = Q/C
= σA ÷ ε₀A/d
= σd/ε₀
Substituting the values of the variables into the equation, we have
V = σd/ε₀
V = 5.0 × 10⁻¹² C/m² × 4.0 × 10⁻³ m/8.854 × 10⁻¹² F/m
V = 20.0 C/m × 10⁻³/8.854 F/m
V = 2.26 × 10⁻³ Volts
V = 2.26 millivolts
So, the potential difference across the parallel plate capacitor is 2.26 millivolts
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Answer:
It will apply the greatest pressure of an area of 1.
Explanation:
To find pressure use the formula P = F/A
P = 100/1
P = 100
D. They are heterotrophs that digest food internally.
Answer:
you could go 12 miles paying $7.80 and $1.75
So in total being $9.55
In 1 hour, the hour hand sweeps across 1/12 of the clock's face. In 40 min, the hour hand travels (40 min)/(60 min) = 2/3 of the path it covers in an hour, so a total of 1/12 × 2/3 = 1/18 of the clock's face. This hand traces out a circle with radius 0.25 m, so in 40 min its tip traces out 1/18 of this circle's radius, or
1/18 × 2<em>π</em> (0.25 m) ≈ 0.087 m
The minute hand traverses (40 min)/(60 min) = 2/3 of the clock's face, so it traces out 2/3 of the circumference of a circle with radius 0.31 m:
2/3 × 2<em>π</em> (0.31 m) ≈ 1.3 m
The second hand completes 1 revolution each minute, so in 40 min it would fully trace the circumference of a circle with radius 0.34 m a total of 40 times, so it covers a distance of
40 × 2<em>π</em> (0.34 m) ≈ 85 m
