An object moving with constant velocity
<span>d. The parallaxes beyond a few thousand light years are
too small to be measured with common instruments.
I'm not sure that parallax can even be used out to a few
thousand light years.
The NEAREST star to Earth has the BIGGEST parallax.
The star is Alpha Centauri. It's only 4 light years away
from us, and its parallax is 0.000206 of a degree !
I have no idea how astronomers can measure angles
so small ... and that's the BIGGEST parallax angle of
ANY star.</span>
Hi there!
Recall the equation for centripetal force;
m = mass (kg)
v = velocity (m/s)
r = radius (m)
We are not given the mass directly, we are given the object's weight.
Now, we can plug this value along with the given velocity and radius to solve:
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
Learn more about potential difference across parallel plate capacitor here:
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