Question

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

Answer 1

Answer:

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