Question

A honey bee of mass 135 mg has accumulated a static charge of +1.7 pC. The bee is returning to her hive by following the path shown in the figure. Because Earth has a naturally occurring electric field near ground level of around 100 V/m pointing vertically downward, the bee experiences an electric force as she flies.

What is the change in the bee’s electric potential energy, Δ?electric, as she flies from point A (0 m,0.5 m) to point B (5.3 m,3.5 m) ?

Compute the ratio of the bee’s change in electric potential energy to her change in gravitational potential energy, Δ?electric/Δ?grav.

A honey bee of mass 135 mg has accumulated a static charge of +1.7 pC. The bee is returning to her hive by following the path shown in the figure. Because Earth has a naturally occurring electric field near ground level of around 100 V/m pointing vertically downward, the bee experiences an electric force as she flies. A. What is the change in the bee's electric potential energy, AUelectrias she flies from point A (O m, 0.5 m) to point B (5.3 m, 3.5 m)? Δ Uelectrie Compute the ratio of the bee's change in electric potential energy to her change in gravitational potential energy, AUclectiAUgrav Δ Uelectrie Ugrav-

Answer 1

Concept: here we use the relation between field and potential to find the change in potential energy. Then we find the ratio of two potential energies, ***************************************************************************************************
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Concept: here we use the relation between field and potential to find the change in potential energy. Then we find the ratio of two potential energies, Delta U_electric = -W_electric = -F vector d vector = -q E vector d vector = -(1.7 pC) (-100 v/m j^-) (3.5 - 0.5 m) j^- = 5.1 times 10^-10 J Delta U_grav = mg h = (135 times 10^-6 kg) (9.8 m/s^2) (3.5 m - 0.5 m) = 3.97 times 10^-3 J So, Delta U_electric/Delta U_gravity = 1.28 times 10^-7