Question

A 0.10 g honeybee acquires a charge of +23 pC while flying.
a. The electric field near the surface of the earth is typically 100 N/C, directed downward. 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?

Answer 1

Answer:

a) $\frac{F}{w} =2.347\times 10^{-6}\ N$

b) $E=4.2609\times 10^7\ N.C^{-1}$ parallel to the earth surface.

• In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.

Explanation:

Given:

mass of the bee, $m=10^{-4}\ kg$

charge acquired by the bee, $q_2=23\times 10^{-12}\ C$

a.

Electrical field near the earth surface, $E=100\ N.C^{-1}$

Now the electric force on the bee:

we know:

$F=\frac{1}{4\pi.\epsilon_0} \times \frac{q_1.q_2}{r^2}$

$F=E.q_2$

$F=100\times 23\times 10^{-12}$

$F=23\times 10^{-10}\ N$

The weight of the bee:

$w=m.g$

$w=10^{-4}\times 9.8$

$w=9.8\times10^{-4}\ N$

Therefore the ratio :

$\frac{F}{w} =\frac{23\times 10^{-10}}{9.8\times10^{-4}}$

$\frac{F}{w} =2.347\times 10^{-6}\ N$

b.

The condition for the bee to hang is its weight must get balanced by the electric force acing equally in the opposite direction.

So,

$F=9.8\times10^{-4}\ N$

$E.q_2=9.8\times10^{-4}\ N$

$E\times 23\times 10^{-12}=9.8\times10^{-4}\ N$

$E=4.2609\times 10^7\ N.C^{-1}$ parallel to the earth surface.

• In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.

Answer 2

The ratio is $2.34*10^{-6}$ and the electric field required so that the bee hangs in air is $4.2*10^7N/C$ upwards

Given that the mass of bee m = 0.1g = $10^{-4} kg$

Charge acquired by the bee q = +23 pC = $23*10^{-12}C$

Electric field = 100N/C downwards

(a) The force F on a charge in electric field E is given by:

F = qE

F =   $23*10^{-12}C$ × 100 N/C

F = 23×$10^{-10}$ N downwards force is experienced by the bee.

Now, the weight of the bee is

W = mg

W= $10^{-4} kg$ × 9.8 m/$s^2$

W = 9.8 × $10^{-4}$ N is the weight of the bee.

The ratio of F and W:

$\frac{F}{W}= 2.34*10^{-6}$

(b) To balance the weight of the bee, the electric field must be directed outwards from the surface of the earth to generate force opposite to the weight which is directed downwards, towards the center of the earth

in this condition:

$F^{'} =W\\\\qE^{'} = mg\\\\E^{'} = mg/q\\\\E^{'} = 10^{-4}*9.8/23*10^{-12}\\\\E^{'}=4.2*10^7N/C$directed outwards the surface of the earth.

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