Question

Honeybees acquire a charge while flying due to friction with the air. A $100mg$bee with a charge of $+23pC$experiences an electric force in the earth’s electric field, which is typically $100N/C$, directed downward.

a. What is the ratio of the electric force on the bee to the bee’s weight?

b. What electric field strength and direction would allow the bee to hang suspended in the air?

Short answer

a. The ratio of the electric force on the bee to the bee’s weight is $\mathbf{2}\mathbf{.}\mathbf{34}\mathbf{}\mathit{x}\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{6}}$

b. The electric field strength and direction that would allow the bee to hang suspended in the air is$\mathbf{4}\mathbf{.}\mathbf{26}\mathbf{}\mathit{x}\mathbf{}{\mathbf{10}}^{\mathbf{7}}\mathbf{}\mathit{N}\mathbf{/}\mathit{C}$

Step-by-step solution

Given information and formula used

Given :

Bee's weight : $100mg$

Bee's charge :$+23pC$

Electric force in the earth’s electric field : $100N/C$, directed downward.

Theory used :

Electric Force produced by Electric field is : $F_E=E·q$

Weight is given by :$W=m·g$

Calculating the ratio of the electric force on the bee to the bee’s weight 

(a) We must determine the force on the bee in relation to the electric weight of the bee.

To begin, we must calculate the electric force generated by the given $E$-field :

$$\begin{gathered} F_E=100N/C\times 23\times {10}^{-12}C \\ =2.3x{10}^{-9}N \end{gathered}$$

The weight of the bee is :

role="math" localid="1649091524292" $$\begin{gathered} W=m·g \\ =0.0001kgx9.8m/s^2 \\ =9.8{10}^{-4}N \end{gathered}$$

Finally, calculating the ratio :

$$\begin{gathered} \frac{F_E}{W}=\frac{2.3\times {10}^{-9}N}{9.8\times {10}^{-4}N} \\ =\boxed{\mathbf{2}\mathbf{.}\mathbf{34}\mathbf{}\mathit{x}\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{6}}} \end{gathered}$$

Calculating the electric field strength and direction

(b) We need to figure out what electric field strength and direction will allow the bee to float in the air.

Because both $F\mathrm{and}W$are downward forces, we introduce a new electric field which produces a new electric force, $F\text{'}$as follows:

$$\begin{gathered} \frac{F\text{'}}{q}=\frac{F_E+W}{q} \\ \Rightarrow E\text{'}=\frac{F_E+W}{q}=\frac{2.3\times {10}^{-9}N+9.8\times {10}^{-4}N}{23\times {10}^{-12}C} \\ =\boxed{\mathbf{4}\mathbf{.}\mathbf{26}\mathbf{}\mathit{x}\mathbf{}{\mathbf{10}}^{\mathbf{7}}\mathbf{}\mathit{N}\mathbf{/}\mathit{C}} \end{gathered}$$