Honeybees accumulate charge as they fly, and they transfer charge to the flowers they visit. Honeybees are able to sense electric fields; tests show that they can detect a change in field as small as 0.77 N/C. Honeybees seem to use this sense to determine the charges on flowers in order to detect whether or not a flower has been recently visited, so they can plan their foraging accordingly. As a check on this idea, let's do a quick calculation using typical numbers for charges on flowers.
Part A If a bee is at a distance of 20 cm, can it detect the difference between flowers that have a 24 pC charge and a 32 pC charge

We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference

Explanation:

For this exercise let's use the electric field expression

E = k q / r²

where k is the Coulomb constant that is equal to 9 109 N m² /C², q the charge and r the distance to the point of interest positive test charge, in this case the distance to the bee

let's calculate the field for each charge

Q = 24 pC = 24 10⁻¹² C

E₁ = 9 10⁹ 24 10⁻¹² / 0.20²

E₁ = 5.4 N / C

Q = 32 pC = 32 10⁻¹² C

E₂ = 9 10⁹ 32 10⁻¹² / 0.2²

E₂ = 7.2 N / C

let's find the difference between these two fields

ΔE = E₂ -E₁

ΔE = 7.2 - 5.4

ΔE = 1.8 N / C

the minimum detection field is

E_minimum = 0.77 N / C

ΔE> E_minimo

We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference

pristina pristina · Answer 2 · 2022-01-28T12:35:05+01:00

The bee will be able to detect the difference between flowers that have a [tex]24\,pC[/tex] charge and a [tex]32\,pC[/tex] charge as the change in their electric fields is greater than the minimum change a bee can detect.

We know that the formula for determining the electric field is given by;

[tex]E=k\times \frac{q}{r^2}[/tex]

Where, [tex]k=8.98\times10^9\,Nm^2/C^2[/tex]

In case of the first flower, [tex]q_1 = 24\,pC=24\times 10^{-12}\,C[/tex]

and [tex]r=20\,cm=0.2\,m[/tex]

Therefore electric field due to charge [tex]q_1[/tex] is;

[tex]E_1 = k\times\frac{q_1}{r^2} = (8.98\times10^{9} \,Nm^2/C^2) \times\frac{(24\times 10^{-12}\,C)}{(0.2\,m)^2} = 5.388\,N/C[/tex]

In case of the second flower, [tex]q_2 = 32\,pC = 32\times 10^{-12}\,C[/tex],

and [tex]r = 20\,cm = 0.2\,m[/tex]

Therefore electric field due to charge [tex]q_2[/tex] is;
[tex]E_2 = k\times\frac{q_1}{r^2} = (8.98\times10^{9} \,Nm^2/C^2) \times\frac{(32\times 10^{-12}\,C)}{(0.2\,m)^2} = 7.184\,N/C[/tex]

So, the difference between the electric fields of the two flowers is;

[tex]\Delta E = E_2-E_1 = 1.796\, N/C[/tex]

Given that the minimum change a bee can detect is [tex]0.77\,N/C[/tex].

Here, [tex]\Delta E > 0.77\,N/C[/tex]

Therefore we can say that the bee will be able to detect the difference between flowers that have a 24 pC charge and a 32 pC charge.

Learn more about electric fields here:

https://brainly.com/question/4273177