Acharge of 9 µc is on the y axis at 1 cm, and a second charge of −9 µc is on the y axis at −1 cm. find the force on a charge of 7 µc on the x axis at x = 6 cm. the value of the coulomb constant is 8.98755 × 10^9 n*m^2/c^2. answer in n units of n.

The net force on X is Fx=75.4N

The net force on Y is FY=25.17N

Explanation:

This is an electrostatic problem, we can calculate de force applying the formula:

the force because of charge at 1cm on the X axis, will only have an X component of force, so:

For the force because of the charge of the Y axis we have to find the distances usign pitagoras, and the angle:

we can find the angle with:

We now can calculate the force of the X axis because of the second charge:

and for the Force on Y axis:

The net force on X axis is:

Fx=226.48N-151.08N=75.4N

Fy=-25.17N

The magnitude of the resultant force is equal to 0.0216 N

The direction is along the negative y axis

Explanation:

According to the exercise data:

q1 = 9x10^-6 C (0,1)

q2 = -9x10^-6 C (0,-1)

q = 7x10^6 C (6,0)

the distance between load q1 and load q will be equal to:

The same way to calculate the distance between q2 and q:

The force on q due to the load q1, is calculated with the following equation:

The same way to force F2:

F2 = (8.987559x10^9*-9x10^-6*7x10^-6))/(6.08^2)*(cos45i-sin45j) = 0.0108 N(i-j)

The resultant force:

F = F1 + F2 = 0.0108 N *(-2j) = - 0.0216 N j

The magnitude is equal to 0.0216 N

The direction is along the negative y axis

asthenosphere and convection

scattering of light by tiny particles in air to produce the colors we see on sky, is dependent on the following factors:

(i) wavelength of light

(ii) size of the scattering particle  

(iii) distance traveled by light through the atmosphere.

according to rayleigh's law of scattering, the amount of scattering is inversely proportional to the fourth power of its wavelength. shorter the wavelength, greater is its scattering. hence violet/blue light is scattered more than red.

sun's light is a composite light ( visible light) with wavelengths in the range 400 nm (violet) to 700 nm (red). as sunlight passes through our atmosphere, the molecules of air scatter light according to their wavelengths. during sunrise and sunset, the sun's position is below the horizon. hence light from the sun travels a greater distance in the atmosphere when compared to its position a noon.

as sunlight enters the earth's atmosphere, blue light is scattered away and at the end of its journey in the sky, only red and orange wavelengths remain in the beam.

hence at sunrise and sunset, the sky appears reddish orange.