A $0.10g$honeybee acquires a charge of $+23pC$while flying.

a. The earth’s electric field near the surface is typically ($100N/C$, 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?

The force exerted on the charge as a result of its interaction with the electric field at a specific area can be measured. The force on an object is defined by its charge as well as the electric field $\overrightarrow{E}$produced by all other charged particles in the area, and it is given by

Equation $1$

$F_{\mathrm{C}}=qE$

$+23pC$specifies the charge. Convert the unit to Coulomb by multiplying it by two.

$q=(+23\mathrm{pC})\left(\frac{1\times {10}^{-12}\mathrm{C}}{1\mathrm{pC}}\right)=+23\times {10}^{-12}\mathrm{C}$

We can now fill in our $E$and $q$values to get the electric force on the charge by

role="math" localid="1650356308956" $F_{\mathrm{C}}=Eq$

$=(100\mathrm{N}/\mathrm{C})\left(23\times {10}^{-12}\mathrm{C}\right)$

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

The weight equals the mass times the gravitational acceleration, according to Newton's second law.

Equation$2$

$Fg=mg$

To get the weight $F_{\mathrm{G}}$, we plug the values for $m$and $g$into equation ($2$).

$F_{\mathrm{G}}=mg$

$=\left(0.1\times {10}^{-3}\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)$

$=9.8\times {10}^{-4}\mathrm{N}$

We're going to divide the electric force now. The gravitational force is represented by $F$. To find the ratio between them, use $F_{\mathrm{G}}$or the weight.

$\begin{array}{c}\frac{F_{\mathrm{C}}}{F_{\mathrm{G}}}=\frac{23\times {10}^{-10}\mathrm{N}}{9.8\times {10}^{-4}\mathrm{N}}\\ \frac{F_{\mathrm{C}}}{F_{\mathrm{G}}}=2.3\times {10}^{-6}\end{array}$

When the electric force $F$equals the weight, the bee is suspended in the air. As a result, we can obtain an expression for $E$by.

$F_{\mathrm{G}}=F_{\mathrm{C}}$

$F_{\mathrm{G}}=qE$

Equation$3$

$E=F_{\mathrm{G}}/q$

To calculate $E$, we now plug the values of $q$and $F_{\mathrm{G}}$ into equation $3$.

$E=F_{\mathrm{G}}/q$

$=9.8\times {10}^{-4}N/(23\times {10}^{-12}C)$

role="math" localid="1650358569873" $=4.3\times {10}^{7}N/C$

The electric field and the electric force applied on the positive charge have the same direction, hence the electric field is upwards.