What is the capacitance of a parallel place capacitor with plates of area 1.0 cm2 that are separated by a distance of 0.1 mm? assume that strontium titanate is the dielectric.

C = 0.274μF.

The capacitance when a dielectric is present is given by the equation:

C = kε₀(A/d)

Where k is the dielectric constant of the material, ε₀ is a universal constant, A is the area of the capacitor's plates, and d is the distance that separate the plates.

A parallel plates capacitor with plates of area 1.0cm² that are separated by a distance of 0.1mm, the dielectric constant of the stronium titanate is 310 at a temperature of 20 ° C.

K = 310.00, ε₀ = 8.85x10⁻¹²F/m, A = 1.00x10⁻²m², and d = 0.10x10⁻³m. Substituing the values in the equation C = kε₀(A/d):

C = (310.00)(8.85x10⁻¹²F/m)(1.00x10⁻²m²/0.10x10⁻³m)

C = (2.74x10⁻⁹F/m)(100m)

C = 2.74x10⁻⁷F

C = 0.274x10⁻⁶ = 0.274μF

The speed of the coin is 0.2585 m/s when the spring returns to its relaxed position.

Explanation:

The length the spring is extended by = 0.24 - 0.17 = 0.07 m

Stiffness of spring = 0.3 N/m

In order to solve this, we simply need to take convert all the potential energy of the spring into kinetic energy for the coin. This is:

Potential energy = 0.5 * k * x^2

Potential energy = 0.5 * 0.3 * 0.07^2

Potential energy = 0.000735 Joules

Since this is equal to kinetic energy we have:

Kinetic energy = 0.000735

0.5*m*V^2 = 0.000735

0.5 * 0.022 * V^2 = 0.000735

V^2 = 0.06682

V = 0.2585 m/s Or 28.85 cm/s