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:
brainly.com/question/12993474
Nope answer is 500 N downward
draw the freebody diagram to understand
Answer:
1.718 N , attractive
Explanation:
r = 0.66 m, n = 5.7 x 10^13
q1 = 5.7 x 10^13 x 1.6 x 10^-19 = 9.12 x 10^-6 C
q2 = - 5.7 x 10^13 x 1.6 x 10^-19 = - 9.12 x 10^-6 C
F = K q1 q2 / r^2
F = 9 x 10^9 x 9.12 x 10^-6 x 9.12 x 10^-6 / (0.66)^2
F = 1.718 N
As both the charges are opposite in nature, so the force between them is attractive.
Answer:
It is a predator.
Explanation:
They eat other insects so therefore I think they're a predator.
