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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"Frequency" just means "often-ness" ... how often something happens.
It's always expressed as
<em>(number of happenings) / (some period of time) .</em>
I believe the correct answer from the choices listed above is option D. The proportion of carbon-14 in an organism is useful in figuring out the age of that organism after it dies because <span>the proportion of carbon-14 slowly decreases after the death of the organism. Hope this answers the question.</span>
Answer:
1.64 * 10^(-5) m
Explanation:
Parameters given:
Angular separation, θ = 0.018 rad
Wavelength, λ = 589 nm = 5.89 * 10^(-7) m
The angular separation when there are 2 slots is given as
θ = λ/2d
where d = separation between slits
d = λ/2θ
d = (589 * 10^(-9))/(2 * 0.018)
d = 1.64 * 10^(-5) m
Answer:
If an object has a fast velocity, the dots on a ticker tape diagram will be far apart.
