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Measurements show that a honeybee in active flight can acquire an electrostatic charge as great as 93.0 pC
a. how many electrons must be transferred to produce this charge?
b.suppose 2 bees, each with a charge of 93.0pC, are separated by a distance of 1.09cm. Treating the bees as point charges, find the magnitude of the electrostatic force experienced by the bees.
a. number=charge/e where e is the charge on one electron.
b. Use coulombs force equation.
To solve these questions, we'll need to use the concepts of charge and the electrostatic force.
a. To determine the number of electrons required to produce a charge of 93.0 pC, we can use the elementary charge as a reference. The elementary charge, denoted as "e," is approximately equal to -1.602 x 10^-19 coulombs.
To find the number of electrons, we divide the total charge (93.0 pC) by the elementary charge:
Number of electrons = Total charge / Elementary charge
Substituting the values:
Number of electrons = (93.0 x 10^-12 C) / (-1.602 x 10^-19 C)
Using the appropriate units:
Number of electrons = 5.80 x 10^7 electrons
Therefore, approximately 5.80 x 10^7 electrons must be transferred to produce a charge of 93.0 pC.
b. To find the magnitude of the electrostatic force between the two bees, we can use Coulomb's Law. According to Coulomb's Law, the electrostatic force between two point charges is given by the equation:
Force = (k x q1 x q2) / r^2
Where:
- k is the electrostatic constant (approximately equal to 8.99 x 10^9 N m^2 / C^2).
- q1 and q2 are the charges of the two bees.
- r is the distance between the charges.
Substituting the values into the equation:
Force = (8.99 x 10^9 N m^2 / C^2) x (93.0 x 10^-12 C)^2 / (1.09 x 10^-2 m)^2
Calculating the result:
Force ≈ 7.63 x 10^-9 N
Therefore, the magnitude of the electrostatic force experienced by the two bees is approximately 7.63 x 10^-9 Newtons.