Question

A 0.15g honeybee acquires a charge of 21pC while flying. The electric field near the surface of the earth is typically 100N/C , directed downward.

A). What is the ratio of the electric force on the bee to the bee's weight? Fe/W = ?

B). What electric field strength would allow the bee to hang suspended in the air? (in N/C)

C). What would be the necessary electric field direction for the bee to hang suspended in the air? Upward, downward or horizontally directed?

Answer 1

Concepts and reason

The concepts behind in this question are electric force exerted due to electric field, electric field near the surface of the earth, and weight.

Initially calculate the electric force on the bee by using charge acquired by honey bee, and then calculate the weight of the bee exerted by the gravity force. Finally find the ratio of the electric force on the bee to the bee’s weight.

Use the force exerted on a honey bee find the direction of the electric field necessary for the bee to suspend freely in the air.

Fundamentals

The electric force is an attractive or repulsive force between the two charged objects. Here honey bee acquires a charge while flying under the uniform electric field near the surface of the earth, which is directed downward

The expression for the electric force on the bee by using charge acquired by honey bee and electric field on the earth surface is,

$F = Eq$

Here, $F$ is the electric force exerted on the bee, $E$ is the electric field related to earth surface, and $q$ is the charge acquires by the honey bee.

The expression for the bee’s weight in which the gravitational force exerted on it is,

$W = mg$

Here, $W$ is the weight of the honey bee, $m$ is the mass of the honey bee, and $g$ is the acceleration due to gravity.

(A)

The expression for the electric force on the charged honey bee is,

${F_{\rm{e}}} = Eq$

Substitute $100\,{\rm{N/C}}$ for $E$ and $21\,{\rm{pC}}$ for $q$ to find the electric force.

$\begin{array}{c}\\{F_{\rm{e}}} = \left( {100\,{\rm{N/C}}} \right)\left( {21\,{\rm{pC}}} \right)\left( {\frac{{{{10}^{ - 12}}\,{\rm{C}}}}{{1\,{\rm{pC}}}}} \right)\\\\ = 21.0 \times {10^{ - 10}}\,{\rm{N}}\\\end{array}$

The weight of the honey is calculated by the force due to gravity acting on the earth surface is,

$W = mg$

Substitute $0.15\,{\rm{g}}$ for $m$ and $9.8\,{\rm{m/}}{{\rm{s}}^2}$ for $g$ to find weight.

$\begin{array}{c}\\W = \left( {0.15\,{\rm{g}}} \right)\left( {\frac{{0.001\,{\rm{kg}}}}{{1\,{\rm{g}}}}} \right)\left( {9.8\,{\rm{m/}}{{\rm{s}}^2}} \right)\\\\ = 14.7 \times {10^{ - 4}}\,{\rm{N}}\\\end{array}$

Calculate the ratio of the electric force on the bee to the bee’s weight.

$R = \frac{{{F_{\rm{e}}}}}{W}$

Here ${F_e}$ is the electric force exerted on the bee and $W$ is the weight of the honey bee.

Substitute $21.0 \times {10^{ - 10}}\,{\rm{N}}$ for ${F_{\rm{e}}}$ and $14.7 \times {10^{ - 4}}\,{\rm{N}}$ for $W$ , to find the ratio of the electric force to the bee’s weight.

$\begin{array}{c}\\R = \frac{{21.0 \times {{10}^{ - 10}}\,{\rm{N}}}}{{14.7 \times {{10}^{ - 4}}\,{\rm{N}}}}\\\\ = 1.43 \times {10^{ - 6}}\\\end{array}$

(B)

The expression for the electric force for the charged honey while flying is,

${F_{\rm{e}}} = Eq$

The weight of the honey bee in terms of mass and acceleration of gravity is,

$W = mg$

Compare the equations to rewrite the expression for the electric field strength that allows the bee to hang suspended freely in air.

$\begin{array}{c}\\{F_{\rm{e}}} = W\\\\Eq = mg\\\\E = \frac{{mg}}{q}\\\end{array}$

Substitute $0.15\,{\rm{g}}$ for $m$ , $9.8\,{\rm{m/}}{{\rm{s}}^2}$ for $g$ , and $21\,{\rm{pC}}$ for $q$ to find the electric field strength.

$\begin{array}{c}\\E = \frac{{\left( {0.15\,{\rm{g}}} \right)\left( {\frac{{0.001\,{\rm{kg}}}}{{1\,{\rm{g}}}}} \right)\left( {9.8\,{\rm{m/}}{{\rm{s}}^2}} \right)}}{{\left( {21\,{\rm{pC}}} \right)\left( {\frac{{{{10}^{ - 12}}\,{\rm{C}}}}{{1\,{\rm{pC}}}}} \right)}}\\\\ = 7.0 \times {10^7}\,{\rm{N/C}}\\\end{array}$

(C)

Since, the bee is hanging suspended freely in the air so that its direction of electric field should be placed upward direction.

Ans: Part A

The ratio of the electric force on the bee to the bee’s weight is $1.43 \times {10^{ - 6}}$ .