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

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm.   the wire diameter is d = 0.9 mm.   the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a                                          

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2)   in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation:

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more