Respuesta :
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
