You are setting up a physics experiment to measure the electric field of a dipole along its axis. At what distance would you expect the prediction from approximate formula, 2kqs/r^3, to be within 5% of your measurement

y = 1.73 √s

Explanation:

For this question, let's look for the complete formula for the elective field of a dipole and then compare with the approximate formula.

A dipole is two charges of equal magnitude and different sign separated a distance 2, the field on axes at the midpoint is

               E₁ = E₂ = k q² / r²

For distance we use Pythagoras' theorem

              r² = y² + s²

The total electric field is

              E = 2 E₁ cos θ

The field perpendicular to the dipole axis is canceled, let's use trigonometry

            cos θ = s / r

Let's replace

              E = 2 k q² / (y² + s²) a / √(y² + s²)

            E = 2 q s / (y² + s²)^{3/2}

This is the exact formula.

The approximate formula is

            E’= 2 q s / y³

If we relate these two formulas

          E’/ E = (y² + s²).^{3/2}/y³

We see that the error in the distance propagates in an error for the electric field, they ask us that the uncertainty be 5% (er = 0.05)

The approximate formula is the measured value and the exact formula is the actual calculated value, so the relative uncertainty is

       E’= E  (y² + s²).^{3/2} / y³

      ΔE ’= dE’ /dy Δy + dE’/ds Δs + dE’ /dE ΔE

The last term is considered zero since the value is exact

      dE’/ dy =  (y² + s²).^{3/2} (-3y⁻⁴) + y⁻³ 3/2 (y² + s²).^{1/2}  2y

      dE ’/ dy = -3 (y² + s²).^{3/2}/y⁴ - 3y  (y2 + s2).^{1/2} /y³

      dE ’/ ds = 3/2 (y² + s²).^{1/2} 2s/y³

       dE'/ds = 3s (y²+s²).^{1/2} /y³

      ΔE’= E [+3 (y² + s²).^{3/2}/y⁴ + 3y  (y2 + s2).^{1/2} /y³] Δy

                  + [3s (y²+s²).^{1/2} /y³] Δs

     ΔE’/E’ = Δy [3y - 3 / (y² + s²)] + Δs [3s / (y² + s²)]

     ΔE’/E' =  3Δy [(1- / (y² + s²)] + 3Δs  s / (y² + s²)

In general the distance and is measured with a tape measure, large value with an uncertainty of Δy = 0.1 cm and the distance between the charge is measured with a caliper Δs = 0.05 cm

Let's replace the values

           0.05 = 0.1  3[1 – 1/ (y² + s²)] + 0.05 3s /(y² + s²)

This is the formula of the error between the approximate field and the exact field, so that the error is at 0.05, the first term must be eliminated by which  y >> s

          0.05 = 0.05 3s / y²

          1 = 3s / y ²

          y = √3s

          y = 1.73 √s

e

explanation:

there is no change in the electric potential energy as it is in rest on both conditions.

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.