A car is traveling at 118 km/h. What is its speed in mm/h?

If the plate separation is modified after the battery is disconnected, the updated distance between plates is 9.21 mm

If changes are made while the battery remains connected, the new separation becomes 0.11 mm

The capacitance for an air-filled parallel plate capacitor can be expressed as:

In this equation, refers to the permittivity of free space, A stands for the plate area, and D represents the separation distance.

Thus,

.......(1)

Therefore, should the distance between the plates shift from d₁ to d₂, the capacitance ratio in both scenarios can be represented as:

......(2)

Scenario (i)

When the capacitor is fully charged and then disconnected from the battery before adjusting the plate distance, the charge will remain steady while the capacitance varies.

The initial energy E₁ stored in the capacitor can be expressed as:

......(3)

Once the separation changes to d₂, capacitance becomes C₂, but the charge Q remains unchanged.

Thus,

......(4)

By dividing equation (4) by (3),

According to equation (2),

This results in a 3.5 fold increase in energy.

Scenario (2)

If the capacitor is kept connected to the power source, the voltage V across the plates will remain unchanged.

The initial energy is described as

......(5)

The final energy when the plate separation transitions to d₂ can be written as:

.....(6)

Referencing equations (5) and (6)

From equation (2),

Thus, in this particular scenario,

Therefore,

Adjusting plate separation after battery disconnection yields 9.21 mm

If modified while connected, the new separation measures 0.11 mm

Response:

Reasoning:

We will utilize a Gaussian surface that resembles the curved wall of a cylinder, with a radius of 3mm and a length of 1 unit directed parallel to the wire axis.

The charge within this cylinder amounts to 250 x 10⁻⁹ C.

Let E denote the electric field at the curved surface, perpendicular to it.

The total electric flux leaving the curved surface

is calculated as 2π r x 1 x E

or 2 x 3.14 x 3 x 10⁻³ E

According to Gauss's law, the total flux is given by the charge within divided by ε (the charge inside the cylinder being 250 x 10⁻⁹C)

equals 250 x 10⁻⁹ / 2.5 x 8.85 x 10⁻¹²   (where ε = 2.5 ε₀ = 2.5 x 8.85 x 10⁻¹²)

resulting in 11.3 x 10³ weber.

Thus,

2 x 3.14 x 3 x 10⁻³ E = 11.3 x 10³

E =  11.3 x 10³ /  2 x 3.14 x 3 x 10⁻³

=.599 x 10⁶ N /C.

The heat required to raise the temperature of a substance by is represented by

where m stands for the mass of the substance and indicates the specific heat of the substance. In this situation, we possess and , the specific heat of water.
Consequently, we can ascertain the temperature rise :

Initially, the water's temperature was , so the end temperature should be

Thus, the water is expected to be vapor by now.

However, to give a more accurate statement, during the liquid to vapor transition, the heat added to the system is used to break molecular bonds instead of raising the system's temperature. The heat necessary for the phase change from liquid to vapor is expressed as

where denotes the latent heat of vaporization for water.
Nevertheless, the initial heat input of 50 KJ is less than this requirement, indicating there isn't sufficient heat to finish the liquid-vapor transition. Therefore, the water will remain in the liquid-vapor change phase at a temperature of (the temperature at which the phase change begins)