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To solve this problem, Coulomb's law will be applied as follows:
F = k*q1*q2 / r^2 where:
F indicates the force magnitude between the charges
k is a constant = 9.00 * 10^9 N.m^2/C^2
q1 = <span>+2.4 × 10–8 C
q2 = </span><span>+1.8 × 10–6 C
r represents the distance separating the charges = </span><span>0.008 m

By substituting these values, we derive:
F = (9*10^9)(2.4*10^-8)(1.8*10^-6) / (0.008)^2 = 6.075, which rounds to 6.1 Newtons

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The ball was released from a height of 20 meters

Explanation:

The scenario is as follows:

1. A ball drops from the edge of a cliff.

2. Upon reaching the ground, the energy held in its gravitational potential energy transforms entirely into kinetic energy.

   This implies K.E = P.E.

3. The ball impacts the ground at a speed of 20 m/s.

4. The gravitational field strength noted is 10 N/kg.

<pOur goal is to ascertain the height from which the ball was dropped.

<pSince the ball was dropped from a cliff, its initial velocity is 0.<p→ K.E =

where v is the final velocity, is the initial velocity, and m is the mass.

<p→ v = 20 m/s and = 0 m/s.<p→ K.E =

→ K.E =

→ K.E = 200 m joules when the ball strikes the ground.

<p→ P.E = mg h

where g is the gravitational field strength, m is mass, and h signifies height.

<p→ g = 10 N/kg.<p→ P.E = m(10)(h)

→ P.E = 10m h joules.

<p→ P.E = K.E.

→ 10m h = 200 m.

Dividing through by 10m yields:

→ h = 20 meters.

The ball was released from a height of 20 meters.

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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.