Question

Honeybees acquire a charge while flying due to friction with the air. A 120 mg bee with a charge of + 23 pC experiences an electric force in the earth's electric field, which is typically 100 N/C, directed downward.What is the ratio of the electric force on the bee to the bee's weight?F/W = ______What electric field strength would allow the bee to hang suspended in the air?E= _______What electric field direction would allow the bee to hang suspended in the air?Upward? or downward?

Answer 1

Answer:

• $\frac{F}{W}=1.95\times10^{-6}$

• $51304447 \frac{N}{C}$

• Upward

Explanation:

The weight of the bee is:

$W=mg=(120\times10^{-6}kg)(9.81\frac{m}{s^{2}})=1.18\times10^{-3}N$

with m the mass and g the gravity acceleration.

Electric force of the bee is related with the electric field of earth by:

$F_{e}=qE=(23\times10^{-12}C)(100\frac{N}{C})=2.3\times10^{-9}$

with q the charge, E the electric field and Fe the electric force.

So:

$\frac{F}{W}=\frac{2.3\times10^{-9}}{1.18\times10^{-3}}=1.95\times10^{-6}$

Because Newton's first law we should make the net force on it equals cero:

$F+F_e+W=0$

$F=-(F_e+W)=-(2.3\times10^{-9} +1.18\times10^{-3})=-1.1800023\times10^{-3}$

with W the weight, Fe the electric force on the bee due earth's electric field and F the force.

So, the applied electric field should be:

$E_a=\frac{-1.1800023\times10^{-3}}{23\times10^{-12}}=-51304447 \frac{N}{C}$

The negative sign indicates that the electric field should be opposite to earth's electric field, so it should be upward.