Ask Your Teacher In Example 24.6, we found that the electric field of a charged disk approaches that of a charged particle for distances y that are large compared to R, the radius of the disk. To see a numerical instance of this, calculate the magnitude of the electric field a distance y = 2.8 m from a disk of radius R = 2.8 cm that has a total charge of 6.5 µC using the exact formula as follows. (Enter your answer to at least one decimal place.)

E = 7394.6N/C

Explanation:

Please find the attached file for the solution

ι вєℓιєνє ιт ιѕ тнє fιяѕт ¢нσι¢є(α.)

нє ѕнσυℓ∂ ѕмιℓє ρσℓιтℓу αи∂ ѕρєαк тσ тнє ρєяѕσи ωнσ ιѕ ιитєяяυριиg αfтєя тнє ∂ιѕ¢υѕѕισи

нσρє ι ¢συℓ∂ нєℓρ.

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.